solved：9/12
upsolved：10/12
A：solved
B：/
C：solved
D：solved
E：upsolved
F：solved
G：solved
H：solved
I：/
J：solved
K：solved
L：solved
A.
题解：
按高度从高到低，每次将区间中最高点给选出来，然后将左右能连边的区间看成虚点，从最高点向虚点连边，再从虚点向该区间中最高点连边；
这样就变成了一个分层的DAG，奇数层虚点（第1层为超级根），偶数层实点；
这个玩意的结构很类似树，所以我们可以用类似搞树的方法搞这个
然后直接统计每个点的深度和每个子DAG的最深深度
然后询问直接用深度搞就行，特别地需要判一下是否属于同一棵子树
1 #include<bits/stdc++.h> 2 using namespace std; 3 4 const int maxn = 100005; 5 int st[17][maxn], lo[maxn], a[maxn], n, m, base; 6 vector<int> g[maxn]; 7 8 struct node 9 { 10 int val, dep, maxdep; 11 }tr[maxn << 1]; 12 13 int find(int l, int r) 14 { 15 int t = lo[r - l + 1]; 16 return max(st[t][l], st[t][r - (1 << t) + 1]); 17 } 18 19 int makedag(int nd, int l, int r, int dep) 20 { 21 //printf("? %d %d %d\n", l, r, dep); 22 tr[nd].dep = dep; 23 int maxval = find(l, r); 24 int lpos = lower_bound(g[maxval].begin(), g[maxval].end(), l) - g[maxval].begin(); 25 int rpos = upper_bound(g[maxval].begin(), g[maxval].end(), r) - g[maxval].begin() - 1; 26 vector<int> tmp; tmp.clear(); tmp.push_back(l - 1); 27 vector< pair<int, int> > inter; inter.clear(); 28 for(int i = lpos; i <= rpos; ++ i) 29 { 30 int p = g[maxval][i]; 31 tmp.push_back(p); 32 tr[p].dep = dep + 1; 33 //printf("! %d %d %d\n", dep, p, tr[p].dep); 34 } 35 tmp.push_back(r + 1); 36 for(int i = 1; i <= rpos - lpos + 2; i ++) 37 if(tmp[i] - 1 > tmp[i - 1]) 38 { 39 base ++; int t = base; 40 int v = makedag(base, tmp[i - 1] + 1, tmp[i] - 1, dep + 2); 41 inter.push_back(make_pair(t, v)); 42 }else inter.push_back(make_pair(0, 0)); 43 int ret = 0; 44 for(int i = 1; i <= rpos - lpos + 1; i ++) 45 { 46 int p = tmp[i]; 47 tr[p].maxdep = max(inter[i - 1].second, inter[i].second); 48 tr[p].maxdep = max(tr[p].maxdep, tr[p].dep); 49 ret = max(ret, tr[p].maxdep); 50 } 51 return ret; 52 } 53 54 int abs(int x){return x < 0 ? -x : x;} 55 56 int lg[maxn], rg[maxn]; 57 stack<int> s; 58 59 int main() 60 { 61 lo[1] = 0; for(int i = 2; i <= 100000; ++ i) lo[i] = lo[i >> 1] + 1; 62 scanf("%d%d", &n, &m); a[0] = a[n + 1] = n + 100; 63 for(int i = 1; i <= n; ++ i) 64 { 65 scanf("%d", &a[i]); 66 g[a[i]].push_back(i); 67 st[0][i] = a[i]; 68 } 69 for(int i = 1; i <= n; ++ i) g[i].push_back(0), g[i].push_back(n + 1), sort(g[i].begin(), g[i].end()); 70 for(int k = 1; (1 << k) <= n; ++ k) 71 for(int i = 1; i + (1 << k) - 1 <= n; ++ i) 72 st[k][i] = max(st[k - 1][i], st[k - 1][i + (1 << (k - 1))]); 73 base = n + 1; makedag(base, 1, n, 0); 74 while(!s.empty()) s.pop(); s.push(0); 75 for(int i = 1; i <= n; ++ i) 76 { 77 while(a[i] > a[s.top()]) s.pop(); 78 lg[i] = s.top() + 1; 79 s.push(i); 80 } 81 while(!s.empty()) s.pop(); s.push(n + 1); 82 for(int i = n; i >= 1; -- i) 83 { 84 while(a[i] > a[s.top()]) s.pop(); 85 rg[i] = s.top() - 1; 86 s.push(i); 87 } 88 while(m --) 89 { 90 int x, y; scanf("%d%d", &x, &y); 91 if(y == 0)printf("%d\n", (tr[x].maxdep - tr[x].dep) >> 1); 92 else 93 { 94 if(a[x] < a[y]) swap(x, y); 95 if(a[x] == a[y] || y < lg[x] || y > rg[x]) printf("0\n"); 96 else printf("%d\n", abs(tr[x].dep - tr[y].dep) >> 1); 97 } 98 } 99 return 0; 100 }
B.
unsolved
C.
签到题
1 #include<bits/stdc++.h> 2 using namespace std; 3 4 const int maxn = 5005; 5 6 pair<int, int> st[maxn]; 7 int n, m, a[maxn], ans[maxn]; 8 9 int main() 10 { 11 scanf("%d%d", &n, &m); 12 for(int i = 1; i <= n; i ++) 13 { 14 scanf("%d", &a[i]); 15 st[i] = make_pair(a[i], i); 16 } 17 sort(st + 1, st + n + 1); 18 int ret = 0; 19 for(int i = 1; i <= n; ++ i) ans[i] = 0; 20 for(int i = 1; i <= n; ++ i) 21 { 22 int pos = st[i].second; 23 int val = -1; 24 for(int j = max(1, pos - m); j <= min(n, pos + m); ++ j) 25 if(a[j] < a[pos]) val = max(val, ans[j]); 26 ans[pos] = val + 1; 27 ret = max(ret, ans[pos]); 28 } 29 printf("%d\n", ret); 30 return 0; 31 }
D.
签到题
1 #include<bits/stdc++.h> 2 using namespace std; 3 4 const int maxn = 2000005; 5 int T, n, a[maxn]; 6 7 int abs(int x){return x < 0 ? -x : x;} 8 9 int main() 10 { 11 scanf("%d", &T); 12 while(T --) 13 { 14 scanf("%d", &n); 15 for(int i = 1; i <= n; ++ i) scanf("%d", &a[i]); 16 sort(a + 1, a + n + 1); 17 int cnt = 0, now = 0; 18 for(int i = 1; i <= n; ++ i) 19 { 20 if(now == 0 || abs(now - a[i]) > 10) 21 { 22 cnt ++; 23 now = a[i] + 10; 24 } 25 } 26 printf("%d\n", cnt); 27 } 28 return 0; 29 }
E.
题解：
写法不优就很容易写挂的题目。。。
考虑dp[i][j]表示dp到值为i，用了j次机会的最长长度
然后讨论转移就行了
注意枚举数字时候不要枚举值，枚举每一个数字会更好写
1 #include<bits/stdc++.h> 2 #define maxn 305 3 using namespace std; 4 int T,n,cas; 5 int a[maxn]; 6 int dp[maxn][3]; 7 int main() 8 { 9 scanf("%d",&T); 10 while(T--) 11 { 12 ++cas; 13 scanf("%d",&n); 14 memset(a,0,sizeof(a)); 15 memset(dp,0,sizeof(dp)); 16 for(int i=1;i<=n;++i)scanf("%d",&a[i]); 17 sort(a+1,a+n+1); 18 for(int i=1;i<=n;++i) 19 { 20 int t=a[i]+100; 21 for(int j=2;j>=1;--j) 22 { 23 dp[t+1][j]=max(dp[t+1][j-1],max(dp[t][j-1],dp[t-1][j-1]))+1; 24 } 25 for(int j=2;j>=0;--j) 26 { 27 dp[t][j]=max(dp[t][j],max(dp[t-1][j],dp[t-2][j]))+1; 28 } 29 for(int j=2;j>=1;--j) 30 { 31 dp[t-1][j]=max(dp[t-1][j-1],max(dp[t-2][j-1],dp[t-3][j-1]))+1; 32 } 33 } 34 int ans=0; 35 for(int i=1;i<=300;++i) 36 for(int j=0;j<=2;++j) 37 { 38 ans=max(ans,dp[i][j]); 39 } 40 printf("Case %d: %d\n",cas,ans); 41 } 42 }
F.
题解：
同UOJ275，利用Lucas定理转化，将C(n,m)中的n,m写成7进制数，存在某一位n的数位比m小
然后按数位计数就行了
1 #include<cstdio> 2 #define mo 1000000007 3 using namespace std; 4 int T; 5 long long n; 6 long long a[70],qr[70],g[70]; 7 int main() 8 { 9 scanf("%d",&T); 10 for (int oo=1;oo<=T;++oo) 11 { 12 scanf("%lld",&n); 13 a[0]=0; 14 long long tmp=n; 15 while (tmp) 16 a[++a[0]]=tmp%7, 17 tmp/=7; 18 qr[0]=1; 19 g[0]=1; 20 for (int i=1;i<=a[0]+1;++i) 21 qr[i]=qr[i-1]*7LL%mo, 22 g[i]=1LL*g[i-1]*28%mo; 23 long long ans=0,p=0,q=1; 24 for (int i=a[0];i>=1;--i) 25 { 26 for (int j=0;j<=a[i]-1;++j) 27 ans=(ans+1LL*((p*7LL+j)%mo*qr[i-1])%mo*qr[i-1]%mo+(qr[i-1]+1LL)*qr[i-1]%mo*((mo+1)/2)%mo-1LL*q*(j+1)%mo*g[i-1]%mo)%mo; 28 p=(p*7LL+a[i])%mo; 29 q=q*(a[i]+1LL)%mo; 30 } 31 long long cnt=1; 32 for (int i=1;i<=a[0];++i) cnt=cnt*(a[i]+1LL)%mo; 33 printf("Case %d: %lld\n",oo,((ans+n+1-cnt)%mo+mo)%mo); 34 } 35 }
G.
题解：
SCC模板题
1 #include<bits/stdc++.h> 2 #define maxn 205 3 using namespace std; 4 int T,n,m; 5 vector<int> g[maxn],scc[maxn]; 6 stack<int> s; 7 int pre[maxn],low[maxn],belong[maxn]; 8 int t,cnt; 9 void dfs(int u) 10 { 11 pre[u]=low[u]=++t; 12 s.push(u); 13 for(int i=0;i<g[u].size();++i) 14 { 15 int v=g[u][i]; 16 if(!pre[v])dfs(v),low[u]=min(low[u],low[v]); 17 else if(!belong[v])low[u]=min(low[u],pre[v]); 18 } 19 if(low[u]==pre[u]) 20 { 21 ++cnt; 22 while(1) 23 { 24 int x=s.top();s.pop(); 25 belong[x]=cnt; 26 scc[cnt].push_back(x); 27 if(x==u)break; 28 } 29 } 30 } 31 int main() 32 { 33 scanf("%d",&T); 34 while(T--) 35 { 36 scanf("%d%d",&n,&m); 37 for(int i=1;i<=n;++i)g[i].clear(),scc[i].clear(),pre[i]=low[i]=belong[i]=0; 38 t=cnt=0; 39 while(!s.empty())s.pop(); 40 for(int u,v,i=1;i<=m;++i) 41 { 42 scanf("%d%d",&u,&v); 43 u++;v++; 44 g[u].push_back(v); 45 } 46 for(int i=1;i<=n;++i)if(!pre[i])dfs(i); 47 printf("%d\n",cnt); 48 } 49 }
H.
题解：
观察（榜）发现可能有一些奥妙重重的性质
可以直接枚举0~2^21 check
1 #include<bits/stdc++.h> 2 #define ll long long 3 using namespace std; 4 int T; 5 ll A,B,C,NA,NB,NC; 6 int main() 7 { 8 scanf("%d",&T); 9 while(T--) 10 { 11 cin>>NA>>NB>>NC>>A>>B>>C; 12 for(int i=0;i<(1<<21);++i) 13 { 14 ll t=1ll*i*i*i; 15 if(t%NA==A&&t%NB==B&&t%NC==C) 16 { 17 cout<<i<<endl; 18 break; 19 } 20 } 21 } 22 }
I.
unsolved
J.
题解：
考虑利用三次方差因式分解提出一个1e-15，然后单独统计前面的部分
这样会被卡常
忽略高阶的1e-15就能过了
1 #include<bits/stdc++.h> 2 using namespace std; 3 4 const double eps = 1e-15; 5 const double mi = 1.0 / 3.0; 6 double sum = 0; 7 int x, y; 8 9 void pri(int p) 10 { 11 char s[10]; 12 int cc = 0; 13 s[3] = s[2] = s[1] = 48; 14 while(p > 0) 15 { 16 s[++ cc] = 48 + (p % 10); 17 p /= 10; 18 } 19 for(int i = 3; i >= 1; i --) putchar(s[i]); 20 puts(""); 21 } 22 23 int main() 24 { 25 while(scanf("%d%d", &x, &y) == 2 && 1ll * x + 1ll * y > 0) 26 { 27 sum = 0; 28 for(int i = x; i <= y; ++ i) 29 { 30 double p1 = (double)i + eps; 31 double p2 = (double)i; 32 double tmp = 3.0 * pow(p1 * p2, mi); 33 //printf("%d %.5Lf\n", i, (double)1.0 / tmp); 34 sum += (double)1.0 / tmp; 35 } 36 int cnt = 15; 37 while(sum + eps < 1) {cnt ++; sum *= 10.0;} 38 while(sum >= 10 + eps) {cnt --; sum /= 10.0;} 39 printf("%.5lfE-", sum); pri(cnt); 40 } 41 return 0; 42 }
K.
题解：
动态维护第k小，线段树上二分
1 #include<bits/stdc++.h> 2 using namespace std; 3 4 const int maxn = 100005; 5 const int maxm = 1000005; 6 7 struct FastIO 8 { 9 static const int S=5<<20; 10 int wpos;char wbuf[S]; 11 FastIO():wpos(0){} 12 inline int xchar() 13 { 14 static char buf[S]; 15 static int len=0,pos=0; 16 if(pos==len)pos=0,len=fread(buf,1,S,stdin); 17 if(pos==len)return -1; 18 return buf[pos++]; 19 } 20 inline int xuint() 21 { 22 int c=xchar(),x=0; 23 while(~c&&c<=32)c=xchar(); 24 for(;'0'<=c&&c<='9';c=xchar())x=x*10+c-'0'; 25 return x; 26 } 27 }io; 28 29 int vsh[maxn], vs[maxn], vc, vsc, lsh[maxn << 1], ls[maxn << 1], lsc, cc, T, n; 30 struct event 31 { 32 int tp, bg, ed, val; 33 void read() 34 { 35 tp = io.xuint(); 36 //scanf("%d", &tp); 37 bg = io.xuint(); 38 val = io.xuint(); 39 if(tp == 1) ed = io.xuint(); 40 lsh[++ cc] = bg; 41 if(tp == 1)lsh[++ cc] = ed + 1; 42 if(tp == 1) vsh[++ vc] = val; 43 } 44 }ev[maxn]; 45 vector<int> g[maxn << 1]; 46 struct node 47 { 48 int ls, rs, l, r, sum; 49 void clear(int x, int y){ls = rs = sum = 0; l = x; r = y;} 50 }tr[maxm << 1]; 51 52 void mt(int x, int y) 53 { 54 tr[cc].clear(x, y); 55 if(x == y) return; 56 int t = cc, mid = (x + y) >> 1; 57 cc ++; tr[t].ls = cc; mt(x, mid); 58 cc ++; tr[t].rs = cc; mt(mid + 1, y); 59 } 60 61 void add(int rt, int pos, int val) 62 { 63 if(tr[rt].l == tr[rt].r) 64 { 65 tr[rt].sum += val; 66 return; 67 } 68 int mid = (tr[rt].l + tr[rt].r) >> 1; 69 if(pos <= mid) add(tr[rt].ls, pos, val); 70 else add(tr[rt].rs, pos, val); 71 tr[rt].sum = tr[tr[rt].ls].sum + tr[tr[rt].rs].sum; 72 } 73 74 int query(int rt, int las) 75 { 76 //printf("!!! %d %d\n", rt, las); 77 if(tr[rt].l == tr[rt].r) return tr[rt].l; 78 int ls = tr[rt].ls, rs = tr[rt].rs; 79 if(tr[ls].sum >= las) return query(ls, las); 80 else return query(rs, las - tr[ls].sum); 81 } 82 83 int main() 84 { 85 //freopen("K.in", "r", stdin); 86 T = io.xuint(); 87 //scanf("%d", &T); 88 for(int cas = 1; cas <= T; ++ cas) 89 { 90 n = io.xuint(); 91 //scanf("%d", &n); 92 cc = 0; vc = 0; 93 for(int i = 1; i <= (n << 1); ++ i) g[i].clear(); 94 for(int i = 1; i <= n; ++ i) ev[i].read(); 95 sort(lsh + 1, lsh + cc + 1); 96 sort(vsh + 1, vsh + vc + 1); 97 ls[lsc = 1] = lsh[1]; vs[vsc = 1] = vsh[1]; 98 for(int i = 2; i <= cc; ++ i) if(lsh[i] != lsh[i - 1]) ls[++ lsc] = lsh[i]; 99 for(int i = 2; i <= vc; ++ i) if(vsh[i] != vsh[i - 1]) vs[++ vsc] = vsh[i]; 100 for(int i = 1; i <= n; ++ i) 101 if(ev[i].tp == 1) 102 { 103 int pos = lower_bound(ls + 1, ls + lsc + 1, ev[i].ed + 1) - ls; 104 int vp = lower_bound(vs + 1, vs + vsc + 1, ev[i].val) - vs; 105 g[pos].push_back(vp); 106 } 107 cc = 1; mt(1, vsc); 108 printf("Case %d:\n", cas); 109 int xx = 1; 110 for(int i = 1; i <= n; ++ i) 111 { 112 int now = lower_bound(ls + 1, ls + lsc + 1, ev[i].bg) - ls; 113 while(xx <= now) 114 { 115 for(int p : g[xx]) add(1, p, -1); 116 xx ++; 117 } 118 if(ev[i].tp == 1) add(1, lower_bound(vs + 1, vs + vsc + 1, ev[i].val) - vs, 1); 119 if(ev[i].tp == 2) 120 { 121 if(ev[i].val > tr[1].sum) puts("-1"); 122 else printf("%d\n", vs[query(1, ev[i].val)]); 123 } 124 } 125 } 126 return 0; 127 }
L.
题解：
二分边长二维前缀和check
卡读入，要fastio
1 #include<bits/stdc++.h> 2 #define maxn 1005 3 using namespace std; 4 int T; 5 int n,m; 6 int a[maxn][maxn],sum[maxn][maxn]; 7 bool check(int x,int y,int r) 8 { 9 return sum[x+r-1][y+r-1]-sum[x-1][y+r-1]-sum[x+r-1][y-1]+sum[x-1][y-1]<=1; 10 } 11 int solve(int x,int y) 12 { 13 int l=0,r=min(n-x+1,m-y+1),ans=0; 14 while(l<=r) 15 { 16 int mid=(l+r)>>1; 17 if(check(x,y,mid))ans=mid,l=mid+1; 18 else r=mid-1; 19 } 20 return ans; 21 } 22 struct FastIO 23 { 24 static const int S=5<<20; 25 int wpos;char wbuf[S]; 26 FastIO():wpos(0){} 27 inline int xchar() 28 { 29 static char buf[S]; 30 static int len=0,pos=0; 31 if(pos==len)pos=0,len=fread(buf,1,S,stdin); 32 if(pos==len)return -1; 33 return buf[pos++]; 34 } 35 inline int xuint() 36 { 37 int c=xchar(),x=0; 38 while(~c&&c<=32)c=xchar(); 39 for(;'0'<=c&&c<='9';c=xchar())x=x*10+c-'0'; 40 return x; 41 } 42 }io; 43 int main() 44 { 45 T=io.xuint(); 46 while(T--) 47 { 48 n=io.xuint();m=io.xuint(); 49 for(int i=1;i<=n;++i) 50 for(int j=1;j<=m;++j)a[i][j]=io.xuint(); 51 for(int i=1;i<=n;++i) 52 for(int j=1;j<=m;++j)sum[i][j]=sum[i-1][j]+sum[i][j-1]-sum[i-1][j-1]+a[i][j]; 53 int ans=0; 54 for(int i=1;i<=n;++i) 55 for(int j=1;j<=m;++j) 56 { 57 ans=max(ans,solve(i,j)); 58 } 59 printf("%d\n",ans); 60 } 61 }