选拔赛的题解，~~~
题目链接：请点击
A题
素数筛 + 线段树（树状数组）
先用素数筛打表，然后线段树更新，遍历求出值，O（1）查询即可
AC代码：
/*num数组 是把记录 数是否存在 存在即为1。 总共有N个数，如何判断第i+1个数到最后一个 数之间有多少个数小于第i个数呢？不妨假设 有一个区间 [1,N]，只需要判断区间[i+1,N]之 间有多少个数小于第i个数。如果我们把总区间初 始化为0，然后把第i个数之前出现过的数都在相应 的区间把它的值定为1，那么问题就转换成了[i+1,N] 值的总和 */ #include <stdio.h> #include <bits/stdc++.h> #define lson l, m, rt<<1 #define rson m+1, r, rt<<1|1 #define size 1000000 + 7 #define min(a, b) a > b ? b : a typedef long long LL; #define MAXN 1000005 #define MAXL 1299710 LL prime[MAXN]; LL check[MAXL]; LL PP[MAXL]; void init(){ LL tot = 0; memset(check, 0, sizeof(check)); for (LL i = 2; i < MAXL; ++i) { if (!check[i]) { prime[tot++] = i; } for (LL j = 0; j < tot; ++j) { if (i * prime[j] > MAXL) { break; } check[i*prime[j]] = prime[j]; if (i % prime[j] == 0) { break; } } } } LL num[size << 2], x[size]; void pushup(LL rt) { num[rt] = num[rt << 1] + num[rt << 1 | 1]; } void build( LL l, LL r, LL rt) { num[rt] = 0; if( l == r) return; LL m = (l + r) >> 1; build(lson); build(rson); } LL qurey( LL L, LL R, LL l, LL r, LL rt) { if( L <= l && r <= R) return num[rt]; LL m = (l + r) >> 1; LL ans = 0; if(L <= m) ans+=qurey(L, R, lson); if(R > m) ans+=qurey(L, R, rson); return ans; } void updata( LL p, LL l, LL r, LL rt) { if( l == r) { num[rt]++;return; } LL m = ( l + r) >> 1; if( p <= m) updata(p, lson); else updata(p, rson); pushup(rt); } LL a[MAXN + 7]; int main() { LL n; init(); check[1] = 1; for(LL i = 2 ; i <= MAXN; i++ ){ if(!check[i]) check[i] = i; } LL sum = 0; build(1, MAXN , 1); for( LL i= 1; i <= MAXN; i++) { LL sum = 0; //scanf("%lld", &x[i]); sum = qurey(1, check[i], 1, MAXN , 1); a[i] = sum; updata(check[i], 1, MAXN, 1); } LL t; scanf("%lld",&t); while(t--){ LL x; scanf("%lld",&x); printf("%lld\n",a[x]); } return 0; }
B题
并查集
没有蘑菇的建树，求出有多少个没有蘑菇的路，然后求出有蘑菇的路就行
AC代码：
/** /*@author Victor /*language C++ */ //#include <bits/stdc++.h> #include<iostream> #include<algorithm> #include<cstdlib> #include<cstring> #include<cstdio> #include<string> #include<vector> #include<bitset> #include<queue> #include<deque> #include<stack> #include<cmath> #include<list> //#include<map> #include<set> //#define DEBUG #define RI register ll using namespace std; typedef long long ll; //typedef __ll128 lll; const ll N=100000+10; const ll MOD=1e9+7; const double PI = acos(-1.0); const double EXP = 1E-8; const ll INF = 0x3f3f3f3f; #define pii pair<ll,ll> #define pll pair<ll,ll> #define pil pair<ll , ll> #define pli pair<ll,ll> #define pdl pair<double,ll> #define pld pair<ll,double> #define pdd pair<double,double> #define iput(n) scanf("%lld",&n); #define iiput(a,n) scanf("%lld%lld",&a,&n); #define iiiput(a,b,c) scanf("%lld%lld%lld",&a,&b,&c); #define dput(n) scanf("%lf",&n); #define llput(n) scanf("%lld",&n); #define cput(n) scanf("%s",n); #define puti(n) prllf("%lld\n",n); #define putll(n) prllf("%lld\n",n); #define putd(n) prllf("%lfd\n",n); #define _cls(n) memset(n,0,sizeof(n)); #define __cls(n) memset(n,INF,sizeof(n)); //priority_queue <ll,vector<ll>,greater<ll> > Q;//优先队列递增 //priority_queue<ll>Q;//递减 //map<ll,ll>mp; //set<ll>st; //stack<>st; //queue<>Q; /***********************************************/ //加速输入挂 # define IOS ios::sync_with_stdio(false) # define FOR(i,a,n) for(ll i=a; i<=n; ++i) //求二进制中1的个数 //__builtin_popcount(n); //求2^k //#define (ll)Pow(2,k) (1LL<<k) #define to_1(n) __builtin_popcount(n) //树状数组 #define lowbit(x) (x&-x) ll fa[333333]; ll pre[333333]; void init(ll n){ for(ll i = 1;i <= n;i++) fa[i] = i; } ll find(ll x) { if(fa[x] == x) return x; return fa[x] = find(fa[x]); } void merge(ll x,ll y) { ll xi = find(x); ll yi = find(y); if(xi == yi) return; else fa[xi] = yi; return ; } int main(){ ll n; IOS; cin >> n; init(n); for(ll i = 1 ; i < n ; i++){ ll u,v,w; cin >> u >> v >> w; if(w == 1) continue; merge(u,v); } for(ll i = 1; i <= n;i++){ ++pre[find(i)]; } ll sum = 0; for(ll i = 1; i <= n;i++){ if(pre[i]) sum += pre[i] * (n - pre[i]); } cout << sum << endl; return 0; }
C题
N次最短路
AC代码：
/** /*@author Victor /*language C++ */ #include <cmath> #include <cstring> #include <cstdio> #include <vector> #include <string> #include <algorithm> #include <string> #include <map> #include <queue> #include <set> #include <stack> using namespace std; #define MAXN 200010 #define LEN 200010 #define INF 1e9+7 #define MODE 1000000 #define pi acos(-1) #define g 9.8 typedef long long ll; #define lson l,m,rt<<1 #define rson m+1,r,rt<<1|1 struct edge{ ll next,to,w; }; edge G[MAXN*2]; ll root; ll num=0; ll cnt=0; ll head[MAXN<<1]; ll depth[MAXN<<1];//深度 ll first[MAXN<<1];//首次出现编号 ll dir[MAXN<<1];//距离 ll que[MAXN<<1];//队列 ll par[MAXN];//并查集父节点 bool vis[MAXN]; ll dp[MAXN][20]; void init(ll n){for(ll i=0;i<=n;i++)par[i]=i;}//初始化并查集 ll _find(ll x){if(par[x]==x)return x;else return par[x]=_find(par[x]);}//查询并查集祖先 void unite(ll x,ll y){x=_find(x),y=_find(y);if(x==y)return;else par[x]=y;}//合并节点 void add(ll u,ll v,ll w){G[num].w=w;G[num].to=v;G[num].next=head[u];head[u]=num++;}//前向星建图 void dfs(ll u,ll dep)//dfs建图 { vis[u]=true; que[++cnt]=u;first[u]=cnt;depth[cnt]=dep; for(ll k=head[u];k!=-1;k=G[k].next) { ll v=G[k].to,w=G[k].w; if(!vis[v]){ dir[v]=dir[u]+w; dfs(v,dep+1); que[++cnt]=u;depth[cnt]=dep; } } } //ST表求LCA void ST(ll n) { for(ll i=1;i<=n;i++) dp[i][0] = i; for(ll j=1;(1<<j)<=n;j++) { for(ll i=1;i+(1<<j)-1<=n;i++) { ll a = dp[i][j-1] , b = dp[i+(1<<(j-1))][j-1]; dp[i][j] = depth[a]<depth[b]?a:b; } } } //两个节点之间的路径深度最小的就是LCA ll RMQ(ll l,ll r) { ll k=0; while((1<<(k+1))<=r-l+1) k++; ll a = dp[l][k], b = dp[r-(1<<k)+1][k]; //保存的是编号 return depth[a]<depth[b]?a:b; } ll LCA(ll u ,ll v) { ll x = first[u] , y = first[v]; if(x > y) swap(x,y); ll res = RMQ(x,y); return que[res]; } ll n,m,c; int main() { scanf("%lld",&n); m = n - 1; num=0,cnt=0; memset(head,-1,sizeof(head)); memset(vis,0,sizeof(vis)); init(n); for(ll i=0;i<m;i++) { ll u,v,w; scanf("%lld%lld%lld",&u,&v,&w); add(u,v,w); add(v,u,w); unite(u,v); } ll u1,v1,w1; scanf("%lld%lld%lld",&u1,&v1,&w1); for(ll i=1;i<=n;i++) { if(_find(i)==i){ dir[i]=0; dfs(i,1); } } ST(2*n-1); scanf("%lld",&c); while(c--) { ll u,v; scanf("%lld%lld",&u,&v); ll lca=LCA(u,v); ll mid1 = dir[u]+dir[v]-2*dir[lca]; ll lca1=LCA(u,u1); ll lca2 = LCA(v1,v); ll mid2 = dir[u]+dir[u1]-2*dir[lca1] + dir[v]+dir[v1]-2*dir[lca2] + w1; ll lca3=LCA(u,v1); ll lca4 = LCA(u1,v); ll mid3 = dir[u]+dir[v1]-2*dir[lca3] + dir[v]+dir[u1]-2*dir[lca4] + w1; ll mine; if(mid1 > mid2){ mine = mid2; } else mine = mid1; if(mine > mid3){ mine = mid3; } //else mine = mid; printf("%lld\n",mine); } return 0; }
D题
gcd求最简分数
AC代码：
// .*/@@@]]``... // .,@@ ,[[\@@@@]`.. .**.*]]]]/@@@@@@@\]]]]*.**. // *@@ ,OOOO]` .[@@@]]@@@@@@[[[[`* *,[[[\@@@@@@]]`.. .*....*****..... // .=@^ OOOOOOOOO\ @@@[` .[\@@@\]/@@@@@@@@@@@@@@@@@\`. // .=@^,OOOOOOOOO[` [@@ *]]]]]]. =@^. // .@@*=OOOOOO/[` \OOOOOOOOOOO^ =@^. // .@@ /OOOOO/* ,,OOOOOOO^ @@* // .@@.OOOO/. ,OOOOOOO^=@^. // .=@^\OO` =O/OOO @@`. // .=@^ ,OOO*,@/. // *@@` [. @@*. // *@@. @@@^. // .=@^ =@^. // *@@ ,]. .@@* // .=@^ =/ [\@\ .\@^. // ./@` */@\. // *@@ *^@@* // *@@ ]]` ,]]. *^@@. // .,@/ @@@@@@^ =@@@@@@ ,^@@. // .@@ =@@@@@@@ @@@@@@@^ =^@@. // .=@^ @@@@@@^ =@@@@@@ *o^@@* // .=@^ ,ooooo^ [[` [[[[[` ,[[. ./oooo^ /o=@/. // .=@^ ,ooooo^ *oooooo. =o/@@`. // *@@ ** =oo=@/. // .,@\ /oo\@@. // .,@@ ,oo//@/. // .*\@\ ,oo/\@@`. // .,\@@] ]]]]] .]oo[]@@@`. // ..,\@@@@]]]* *@@@OOO@@@@\. .,]]]@@@@@/`*. // .*.*[[[@@@@@/ @@O@@@OO@@@O@@` \@@@[*... // ./@^ =@OOOO@@@@OOOO@@^ *o\@^. // .=@/ ,@@@@@@@@@@@@O@O@@OOO@@^,]@@^ =^@@`. // *@@ ,@@OO@@@@@O@OOO@@@[. *o/@@. // .=@^ /@@O@@@O@@@OOO@@ ]. \o\@^. // .@@ ,]]@@@@@@OO@@@@O@@OOO@@\]]]/@@/ =o/@@* // .,@/ ,[[* \@@OOO@OOOOOOO@@[[[` *o\@@^. // .=@^ \@@OOOOOOOOO@@@\` oo=@\. // .@@. .[@@@@@OOOOOO@@` =o^@@* // *@@ *@@OO@@@@@@ =o^@@. // *@@ ,@@@[ *` =o^@@^. // .*................................................. // /** /*@author Victor /*language C++ */ //#include <bits/stdc++.h> #include<iostream> #include<algorithm> #include<cstdlib> #include<cstring> #include<cstdio> #include<string> #include<vector> #include<bitset> #include<queue> #include<deque> #include<stack> #include<cmath> #include<list> //#include<map> #include<set> //#define DEBUG #define RI register int using namespace std; typedef long long ll; //typedef __int128 lll; const int N=100000+10; const int MOD=1e9+7; const double PI = acos(-1.0); const double EXP = 1E-8; const int INF = 0x3f3f3f3f; #define pii pair<int,int> #define pll pair<ll,ll> #define pil pair<int , ll> #define pli pair<ll,int> #define pdl pair<double,ll> #define pld pair<ll,double> #define pdd pair<double,double> #define iput(n) scanf("%d",&n); #define iiput(a,n) scanf("%d%d",&a,&n); #define iiiput(a,b,c) scanf("%d%d%d",&a,&b,&c); #define dput(n) scanf("%lf",&n); #define llput(n) scanf("%lld",&n); #define cput(n) scanf("%s",n); #define puti(n) printf("%d\n",n); #define putll(n) printf("%lld\n",n); #define putd(n) printf("%lfd\n",n); #define _cls(n) memset(n,0,sizeof(n)); #define __cls(n) memset(n,INF,sizeof(n)); //priority_queue <int,vector<int>,greater<int> > Q;//优先队列递增 //priority_queue<int>Q;//递减 //map<ll,ll>mp; //set<ll>st; //stack<>st; //queue<>Q; /***********************************************/ //加速输入挂 # define IOS ios::sync_with_stdio(false) # define FOR(i,a,n) for(int i=a; i<=n; ++i) //求二进制中1的个数 //__builtin_popcount(n); //求2^k //#define (ll)Pow(2,k) (1LL<<k) #define to_1(n) __builtin_popcount(n) //树状数组 #define lowbit(x) (x&-x) ll gcd(ll a,ll b){ return b?gcd(b,a%b):a; } ll qpow(ll p,ll q){ll f=1;while(q){if(q&1)f=f*p;p=p*p;q>>=1;}return f;} ll exgcd(ll l,ll r,ll &x,ll &y){if(r==0){x=1;y=0;return l;}else{ll d=exgcd(r,l%r,y,x);y-=l/r*x;return d;}} //求a关于m的乘法逆元 ll mod_inverse(ll a,ll m){ll x,y;if(exgcd(a,m,x,y)==1)/*ax+my=1*/return (x%m+m)%m;return -1;/*不存在*/} //快速乘 ll qmul(ll a,ll b,ll m){ll ans=0;ll k=a;ll f=1;/*f是用来存负号的*/if(k<0){f=-1;k=-k;}if(b<0){f*=-1;b=-b;}while(b){if(b&1)ans=(ans+k)%m;k=(k+k)%m;b>>=1;}return ans*f;} //中国剩余定理CRT (x=ai mod mi) ll china(ll n, ll *a,ll *m) {ll M=1,y,x=0,d;for(ll i = 1; i <= n; i++) M *= m[i];for(ll i = 1; i <= n; i++) {ll w = M /m[i]; exgcd(m[i], w, d, y);/*m[i]*d+w*y=1*/ x = (x + y*w*a[i]) % M;}return (x+M)%M;} //筛素数，全局：int cnt,prime[N],p[N]; //void isprime(){cnt = 0;memset(prime,true,sizeof(prime));for(int i=2; i<N; i++){if(prime[i]){p[cnt++] = i;for(int j=i+i; j<N; j+=i)prime[j] = false;}}} //快速求逆元 //void inverse(){inv[1] = 1;for(int i=2;i<N;i++){if(i >= M) break;inv[i] = (M-M/i)*inv[M%i]%M;}} //组合数取模，n和m 10^5时，预处理出逆元和阶乘 /* ll fac[N]={1,1},inv[N]={1,1},f[N]={1,1}; ll C(ll a,ll b){ if(b>a)return 0; return fac[a]*inv[b]%M*inv[a-b]%M; } void init(){//快速计算阶乘的逆元 for(int i=2;i<N;i++){ fac[i]=fac[i-1]*i%M; f[i]=(M-M/i)*f[M%i]%M; inv[i]=inv[i-1]*f[i]%M; } } */ int main(){ int t; cin >> t; while(t--){ ll a,b,c,d; cin >> a >> b >> c >> d; ll sum =( b - a + 1) * (d - c + 1); ll su; if(a > d || c > b) su = 0; else if(a > c) { if(b > d) su = d - a + 1; else su = b - a + 1; } else if( a <= c){ if(b > d) su = d - c + 1; else su = b - c + 1; } if(su == 0) printf("0/1\n"); else { if(sum % su == 0){ printf("1/%lld\n",sum / su); } else { printf("%lld/%lld\n",su / gcd(su,sum) , sum / gcd(su,sum)); } } } return 0; }
E题
暴力最小遍历
AC代码：
/** /*@author Victor /*language C++ */ #include<bits/stdc++.h> using namespace std; typedef long long ll; ll a[1000],b[1000][1000]; int main(){ ll n,k; scanf("%lld%lld",&n,&k); ll min = 0; for(int j = 0 ; j < n ; j++){ for(int i = 0 ; i < k ; i++ ){ scanf("%lld",&a[i]); } sort(a,a+k); for(int z = 0 ; z < k; z++ ) { b[j][z] = a[z]; // cout << a[z] << endl; } } for(int i = 0 ; i < k ; i++){ for(int j = 0 ; j < n - 1 ; j++){ min += abs(b[j][i] - b[j + 1][i]); // cout << min << endl; } } printf("%lld\n",min); return 0; }
F题
枚举每点的最短路求和求最小值
AC代码：
/** /*@author Victor /*language C++ */ #include<iostream> #include<cstdio> #include<vector> #include<cmath> #include<algorithm> #include<map> #include<cstring> #include<string> #include<set> #include<queue> #include<fstream> using namespace std; typedef pair<int,int> PII; const int MAXN=1e4+5; const int INF=0x3f3f3f3f; bool vis[MAXN]; int dist[MAXN],head[MAXN],tot; int pre[MAXN]; struct Edge { int from,to,cost,nxt; Edge(){} Edge(int _from,int _to,int _cost):from(_from),to(_to),cost(_cost){} }e[MAXN*2]; void addedge(int u,int v,int w) { e[tot].from=u;e[tot].to=v;e[tot].cost=w; e[tot].nxt=head[u];head[u]=tot++; } struct qnode { int c,v; qnode(int _c=0,int _v=0):c(_c),v(_v){} bool operator < (const qnode &rhs) const {return c>rhs.c;} }; void Dijkstra(int n,int st)//点的编号从1开始 { memset(vis,false,sizeof(vis)); for(int i=0;i<=n;i++) dist[i]=INF; priority_queue<qnode> pq; while(!pq.empty()) pq.pop(); dist[st]=0; pq.push(qnode(0,st)); qnode frt; while(!pq.empty()) { frt=pq.top(); pq.pop(); int u=frt.v; if(vis[u]) continue; vis[u]=true; for(int i=head[u];i!=-1;i=e[i].nxt) { int to=e[i].to; int cost=e[i].cost; if(dist[to]>dist[u]+cost) { dist[to]=dist[u]+cost; pre[to]=u; pq.push(qnode(dist[to],to)); } } } } int main() { int n,m; while(scanf("%d%d",&n,&m)!=EOF) { if(n==0&&m==0) break; tot=0;memset(head,-1,sizeof(head)); int u,v,w; for(int i=1;i<=m;i++) { scanf("%d%d%d",&u,&v,&w); addedge(u,v,w);addedge(v,u,w); } int st,ed; int sum = 0; int mine = INF; for(int st = 1; st <= n;st++){ Dijkstra(n,st); for(int i = 1 ; i <= n;i++ ) { sum += dist[i]; } if(sum < mine){ ed = st; mine = sum ; } sum = 0; } printf("%d %d\n",mine , ed); } return 0; }
G题
逆序对 + 线段树（归并排序
AC代码：
/** /*@author Victor /*language C++ */ #include <iostream> #include <cstdio> #include <bits/stdc++.h> #define _cls(a) memset(a,0,sizeof(a)) typedef long long ll; using namespace std; ll n, a[200005], tmpA[200005], cnt = 0; ll p[200005]; void merge_sort(ll l, ll r, ll *A) { if (l >= r) return ; ll mid = (l + r) >> 1; merge_sort(l, mid, A); merge_sort(mid + 1, r, A); ll pl = l, pr = mid + 1, tmpp = 0; while(pl <= mid && pr <= r) { if (A[pl] <= A[pr]) tmpA[tmpp++] = A[pl++]; else tmpA[tmpp++] = A[pr++], cnt += mid - pl + 1; } while(pl <= mid) tmpA[tmpp++] = A[pl++]; while(pr <= r) tmpA[tmpp++] = A[pr++]; for (ll i = 0; i < tmpp; i++) A[i + l] = tmpA[i]; } int main() { ll t; scanf("%lld",&t); while(t--){ _cls(p); _cls(a); _cls(tmpA); cnt = 0; scanf("%lld", &n); for (ll i = 1; i <= n; i++){ ll x; scanf("%lld", &x); p[x] = i; } for (ll i = 1; i <= n; i++) { ll x; scanf("%lld", &x); a[i] = p[x]; } merge_sort(1, n, a); printf("%lld\n", cnt); } return 0; }
H题
二分 + 前缀和
AC代码：
/** /*@author Victor /*language C++ */ #include<bits/stdc++.h> using namespace std; typedef long long ll; struct name { ll w, v; }; struct name1 { ll a, b; }; ll s[422222]; ll o[422222]; name p[400000 + 10]; name1 p1[400000 + 10]; int main(){ ll n,m ,S; scanf("%lld%lld%lld",&n,&m,&S); for(ll i = 1 ; i <= n;i ++ ) { ll w,v; scanf("%lld%lld",&p[i].w,&p[i].v); } ll l = 1, r = 10000000; ll W ; for(ll i = 1 ; i <= m; i++){ scanf("%lld %lld",&p1[i].a,&p1[i].b); } ll sum = 2000000000000; for(ll k = 1 ; k <= 100; k++){ W = (l + r) / 2; memset(s,0,sizeof(s)); memset(o,0,sizeof(o)); for(ll j = 1; j <= n; j++ ){ if(W <= p[j].w){ s[j] = s[j - 1] + p[j].v; o[j] = 1 + o[j - 1]; } else { s[j] = s[j - 1] ; o[j] = o[j - 1] ; } } ll mine = 0; ll H = 0; // ll sum = 2000000; for(ll i = 1; i <= m ; i++ ){ H += (s[p1[i].b] - s[p1[i].a - 1]) * (o[p1[i].b] - o[p1[i].a - 1]); } if(H < S) { r = W ; } else l = W; mine = abs(H - S); if(sum > mine) sum = mine; } printf("%lld\n",sum); return 0; }
I题
完全背包 + 快速幂
AC代码：
/** /*@author Victor /*language C++ */ #include<bits/stdc++.h> using namespace std; typedef long long ll; ll t; ll dp[200000 + 7]; ll qpow(ll a,ll b,ll m){ ll ans=1; ll k=a; while(b){ if(b&1)ans=ans*k%m; k=k*k%m; b>>=1; } return ans; }int main(){ scanf("%lld",&t); while(t -- ){ memset(dp,0,sizeof(dp)); ll n, m; dp[1] = 1; // dp[0] = 1; scanf("%lld%lld",&n,&m); if(m==1) dp[n] = qpow(2,n,100000000 + 7); else{ for(int i = 0;i < m;i++){ dp[i] = 1; } for(int i = m ; i <= n;i++){ dp[i] += dp[i - 1] + dp[i - m]; dp[i] %= (100000000 + 7); } } // for(int i = 0 ; i <= n ; i++ ){ // printf("%d\n",dp[i]); // } //printf("%d",dp[n]); printf("%lld\n",dp[n] % (100000000 + 7)); } return 0; }
J题
线段树
AC代码：
/** /*@author Victor /*language C++ */ #include <stdio.h> #include <iostream> #include <algorithm> #include <math.h> #include <string.h> #include <vector> #define LL long long using namespace std; LL INF = 0x3f3f3f3f; const LL MAX = 1000000 + 50; LL Sum[MAX << 2]; LL Add[MAX << 2]; LL A[MAX]; //PushUp函数更新节点信息 ，这里是求和 void PushUp(LL rt){ Sum[rt] = Sum[rt<<1] + Sum[rt<< 1|1]; //prLLf("888sum[%ll] = %ll\n", rt, Sum[rt]); } //PushDown下推标记 void PushDown(LL rt,LL ln,LL rn){ //ln,rn为左子树，右子树的数字数量。 if(Add[rt]){ //下推标记 Add[rt<<1]+=Add[rt]; Add[rt<<1|1]+=Add[rt]; //修改子节点的Sum使之与对应的Add相对应 Sum[rt<<1]+=Add[rt]*ln; Sum[rt<<1|1]+=Add[rt]*rn; //清除本节点标记 Add[rt]=0; } } //Build函数建树 void Build(LL l,LL r,LL rt){ //l,r表示当前节点区间，rt表示当前节点编号 //prLLf("l = %ll r = %ll\n", l, r); if(l==r) {//若到达叶节点 Sum[rt] = A[l];//储存数组值 //prLLf("A[%ll] = %lld\n", l, A[l]); //prLLf("sum[%ll] = %ll\n", rt, Sum[rt]); return; } LL m=(l+r)>>1; //左右递归 Build(l,m,rt<<1); Build(m+1,r,rt<<1|1); //更新信息 PushUp(rt); //prLLf("888sum[%ll] = %ll\n", rt, Sum[rt]); } //点修改 A[L] + C void Update(LL L,LL C,LL l,LL r,LL rt){//l,r表示当前节点区间，rt表示当前节点编号 if(l==r){//到叶节点，修改 Sum[rt] = C; //prLLf("sum[%ll] = %ll\n", rt, Sum[rt]); return; } LL m=(l+r)>>1; PushDown(rt,m-l+1,r-m);//下推标记 //根据条件判断往左子树调用还是往右 if(L <= m) Update(L,C,l,m,rt<<1); else Update(L,C,m+1,r,rt<<1|1); PushUp(rt);//子节点更新了，所以本节点也需要更新信息 } //区间修改 void UpdateRange(LL L,LL R,LL C,LL l,LL r,LL rt){//L,R表示操作区间，l,r表示当前节点区间，rt表示当前节点编号 if(L <= l && r <= R){//如果本区间完全在操作区间[L,R]以内 Sum[rt]+=C*(r-l+1);//更新数字和，向上保持正确 Add[rt]+=C;//增加Add标记，表示本区间的Sum正确，子区间的Sum仍需要根据Add的值来调整 return ; } LL m=(l+r)>>1; PushDown(rt,m-l+1,r-m);//下推标记 //这里判断左右子树跟[L,R]有无交集，有交集才递归 if(L <= m) UpdateRange(L,R,C,l,m,rt<<1); if(R > m) UpdateRange(L,R,C,m+1,r,rt<<1|1); PushUp(rt);//更新本节点信息 } //区间查询， 这里是求和 LL QueryRange(LL L,LL R,LL l,LL r,LL rt){//L,R表示操作区间，l,r表示当前节点区间，rt表示当前节点编号 if(L <= l && r <= R){ //在区间内，直接返回 return Sum[rt]; } LL m=(l+r)>>1; //下推标记，否则Sum可能不正确 PushDown(rt,m-l+1,r-m); //累计答案 LL ANS= 0; if(L <= m) ANS += QueryRange(L,R,l,m,rt<<1); if(R > m) ANS += QueryRange(L,R,m+1,r,rt<<1|1); return ANS; } LL Query(LL l, LL r, LL rt, LL k){ if(l == r){ return l; } LL m = (l + r) >> 1; if(Sum[(rt << 1)] >= k){ return Query(l, m, rt << 1, k); } else{ return Query(m + 1, r, rt << 1 | 1, k - Sum[rt << 1]); } } //建树 Build(1,n,1); //点修改 Update(L,C,1,n,1); //区间修改 UpdateRange(L,R,C,1,n,1); //区间查询 LL ANS=Query(L,R,1,n,1); int main(int argc, char const *argv[]) { LL n, q; scanf("%lld%lld", &n, &q); for(LL i = 1; i <= n; i++){ scanf("%lld", &A[i]); } Build(1, n, 1); while(q--){ char op[3]; LL a , b , c; scanf("%s",op); if (op[1] == '1') { scanf("%lld%lld%lld",&a,&b,&c); UpdateRange(a,b,c,1,n,1); } else if(op[1] == '2'){ scanf("%lld%lld%lld",&a,&b,&c); UpdateRange(a , b , -c , 1 , n , 1); } else if(op[1] == '3'){ scanf("%lld%lld",&a,&c); Update(a,c,1,n,1); } else if(op[1] == '4'){ scanf("%lld%lld",&a,&b); printf("%lld\n",QueryRange(a , b , 1 , n , 1)); } } return 0; }