四题就可以当大爷了，一共才六道题。
A、你有一个角色，a点力量b点智力，你现在有c点经验值，1点经验值可以把力量或智力增加1
你现在需要用掉所有的经验值，问你有多少种使用经验值的方案，使得最后力量大于智力
首先如果力量一开始小于智力，先消耗经验值增加力量直到力量大于智力
然后考虑方程组：
(a+x)>(b+y)
(x+y)=c
x和y分别是增加的力量和智力
稍微改动一下：
(a+x)>=(b+y+1)
这样可以求的最小的x满足上述方程组
那么得到2x>=c+b-a+1
注意一下(c+b-a+1)不一定是偶数，所以实际上x=((c+b-a+1)+1)/2
即向上取整。
最后解出y=c-x，总方案数就是[0,y]这个区间中的数字个数，每一个y都可以满足条件
似乎做完了，但是有个坑点需要特判：
如果(b+c)<a，说明不管怎么加一定满足条件，此时应当输出c+1
代码：
#include <bits/stdc++.h> #define int long long #define sc(a) scanf("%lld",&a) #define scc(a,b) scanf("%lld %lld",&a,&b) #define sccc(a,b,c) scanf("%lld %lld %lld",&a,&b,&c) #define schar(a) scanf("%c",&a) #define pr(a) printf("%lld",a) #define fo(i,a,b) for(int i=a;i<b;++i) #define re(i,a,b) for(int i=a;i<=b;++i) #define rfo(i,a,b) for(int i=a;i>b;--i) #define rre(i,a,b) for(int i=a;i>=b;--i) #define prn() printf("\n") #define prs() printf(" ") #define mkp make_pair #define pii pair<int,int> #define pub(a) push_back(a) #define pob() pop_back() #define puf(a) push_front(a) #define pof() pop_front() #define fst first #define snd second #define frt front() #define bak back() #define mem0(a) memset(a,0,sizeof(a)) #define memmx(a) memset(a,0x3f3f,sizeof(a)) #define memmn(a) memset(a,-0x3f3f,sizeof(a)) #define debug #define db double #define yyes cout<<"YES"<<endl; #define nno cout<<"NO"<<endl; using namespace std; typedef vector<int> vei; typedef vector<pii> vep; typedef map<int,int> mpii; typedef map<char,int> mpci; typedef map<string,int> mpsi; typedef deque<int> deqi; typedef deque<char> deqc; typedef priority_queue<int> mxpq; typedef priority_queue<int,vector<int>,greater<int> > mnpq; typedef priority_queue<pii> mxpqii; typedef priority_queue<pii,vector<pii>,greater<pii> > mnpqii; const int maxn=500005; const int inf=0x3f3f3f3f3f3f3f3f; const int MOD=100000007; const db eps=1e-10; int qpow(int a,int b){int tmp=a%MOD,ans=1;while(b){if(b&1){ans*=tmp,ans%=MOD;}tmp*=tmp,tmp%=MOD,b>>=1;}return ans;} int lowbit(int x){return x&-x;} int max(int a,int b){return a>b?a:b;} int min(int a,int b){return a<b?a:b;} int mmax(int a,int b,int c){return max(a,max(b,c));} int mmin(int a,int b,int c){return min(a,min(b,c));} void mod(int &a){a+=MOD;a%=MOD;} bool chk(int now){} int half(int l,int r){while(l<=r){int m=(l+r)/2;if(chk(m))r=m-1;else l=m+1;}return l;} int ll(int p){return p<<1;} int rr(int p){return p<<1|1;} int mm(int l,int r){return (l+r)/2;} int lg(int x){if(x==0) return 1;return (int)log2(x)+1;} bool smleql(db a,db b){if(a<b||fabs(a-b)<=eps)return true;return false;} db len(db a,db b,db c,db d){return sqrt((a-c)*(a-c)+(b-d)*(b-d));} bool isp(int x){if(x==1)return false;if(x==2)return true;for(int i=2;i*i<=x;++i)if(x%i==0)return false;return true;} int t,a,b,c; signed main(){ ios_base::sync_with_stdio(0); cin.tie(0),cout.tie(0); cin>>t; while(t--){ cin>>a>>b>>c; if(a<=b){ int k=min(c,b+1-a); c-=k; a+=k; } if(c==0&&a>b){ cout<<1<<endl; continue; } if(c==0&&a<=b){ cout<<0<<endl; continue; } if(a>(b+c)){ cout<<c+1<<endl; continue; } int x=(c+b-a+2)/2; int y=c-x; if(a+x==b+y){x++,y--;} cout<<y+1<<endl; } return 0; }
B、你有n种攻击，怪物有x个头
每种攻击可以令x-=d[i]，但是如果此时x>0，那么怪物会回复，即使得x+=h[i]
d[i]和h[i]都是给定的，求杀死怪物的最少次数，不能杀死则输出-1
思路很明显，找d[i]-h[i]最大的攻击，疯狂打怪，最后用d[i]最大的攻击终结他
如果没有d[i]-h[i]>0的攻击并且就连最大的d[i]也一击打不死怪就输出-1
写本文时候还在open hacking，很多人B没了，听说没判一击必杀？
代码：
#include <bits/stdc++.h> #define int long long #define sc(a) scanf("%lld",&a) #define scc(a,b) scanf("%lld %lld",&a,&b) #define sccc(a,b,c) scanf("%lld %lld %lld",&a,&b,&c) #define schar(a) scanf("%c",&a) #define pr(a) printf("%lld",a) #define fo(i,a,b) for(int i=a;i<b;++i) #define re(i,a,b) for(int i=a;i<=b;++i) #define rfo(i,a,b) for(int i=a;i>b;--i) #define rre(i,a,b) for(int i=a;i>=b;--i) #define prn() printf("\n") #define prs() printf(" ") #define mkp make_pair #define pii pair<int,int> #define pub(a) push_back(a) #define pob() pop_back() #define puf(a) push_front(a) #define pof() pop_front() #define fst first #define snd second #define frt front() #define bak back() #define mem0(a) memset(a,0,sizeof(a)) #define memmx(a) memset(a,0x3f3f,sizeof(a)) #define memmn(a) memset(a,-0x3f3f,sizeof(a)) #define debug #define db double #define yyes cout<<"YES"<<endl; #define nno cout<<"NO"<<endl; using namespace std; typedef vector<int> vei; typedef vector<pii> vep; typedef map<int,int> mpii; typedef map<char,int> mpci; typedef map<string,int> mpsi; typedef deque<int> deqi; typedef deque<char> deqc; typedef priority_queue<int> mxpq; typedef priority_queue<int,vector<int>,greater<int> > mnpq; typedef priority_queue<pii> mxpqii; typedef priority_queue<pii,vector<pii>,greater<pii> > mnpqii; const int maxn=500005; const int inf=0x3f3f3f3f3f3f3f3f; const int MOD=100000007; const db eps=1e-10; int qpow(int a,int b){int tmp=a%MOD,ans=1;while(b){if(b&1){ans*=tmp,ans%=MOD;}tmp*=tmp,tmp%=MOD,b>>=1;}return ans;} int lowbit(int x){return x&-x;} int max(int a,int b){return a>b?a:b;} int min(int a,int b){return a<b?a:b;} int mmax(int a,int b,int c){return max(a,max(b,c));} int mmin(int a,int b,int c){return min(a,min(b,c));} void mod(int &a){a+=MOD;a%=MOD;} bool chk(int now){} int half(int l,int r){while(l<=r){int m=(l+r)/2;if(chk(m))r=m-1;else l=m+1;}return l;} int ll(int p){return p<<1;} int rr(int p){return p<<1|1;} int mm(int l,int r){return (l+r)/2;} int lg(int x){if(x==0) return 1;return (int)log2(x)+1;} bool smleql(db a,db b){if(a<b||fabs(a-b)<=eps)return true;return false;} db len(db a,db b,db c,db d){return sqrt((a-c)*(a-c)+(b-d)*(b-d));} bool isp(int x){if(x==1)return false;if(x==2)return true;for(int i=2;i*i<=x;++i)if(x%i==0)return false;return true;} int t,n,x; int d[maxn],h[maxn]; signed main(){ ios_base::sync_with_stdio(0); cin.tie(0),cout.tie(0); cin>>t; while(t--){ cin>>n>>x; re(i,1,n) cin>>d[i]>>h[i]; int p1=-1,mx1=0; int p2=-1,mx2=-inf; re(i,1,n){ if(d[i]-h[i]>mx1){ mx1=d[i]-h[i]; p1=i; } if(d[i]>mx2){ mx2=d[i]; p2=i; } } if(p1==-1&&mx2<x) cout<<-1<<endl; else{ x-=mx2; int ans; if(x%mx1==0) ans=x/mx1; else ans=x/mx1+1; cout<<ans+1<<endl; } } return 0; }
C、给定一个01串，一个好01串的定义是这个01串所代表的二进制数字刚好等于它的长度，允许前导零
问这个01串当中有几个好子串
先不考虑前导零，因为整个串的长度不超过二十万，那么实际上好子串的最大长度也不过18，那么暴力枚举就变得可行了
首先枚举长度，设为len，找到所有的s[i]=1的位置，算出子串[i , i+len-1]所代表的数字cal，暴力算即可
那么这个子串能对答案产生贡献的条件有三个，必须同时满足：
1、cal大于等于len
2、i左侧的位置数一定要大于等于cal-len，这样才有希望通过补上前导零使得cal等于len
3、i-1前面的前导零个数大于等于cal-len，这说明前面有足够的前导零，如果i-1前面有任何一个1都不会满足该条件
代码：
#include <bits/stdc++.h> #define int long long #define sc(a) scanf("%lld",&a) #define scc(a,b) scanf("%lld %lld",&a,&b) #define sccc(a,b,c) scanf("%lld %lld %lld",&a,&b,&c) #define schar(a) scanf("%c",&a) #define pr(a) printf("%lld",a) #define fo(i,a,b) for(int i=a;i<b;++i) #define re(i,a,b) for(int i=a;i<=b;++i) #define rfo(i,a,b) for(int i=a;i>b;--i) #define rre(i,a,b) for(int i=a;i>=b;--i) #define prn() printf("\n") #define prs() printf(" ") #define mkp make_pair #define pii pair<int,int> #define pub(a) push_back(a) #define pob() pop_back() #define puf(a) push_front(a) #define pof() pop_front() #define fst first #define snd second #define frt front() #define bak back() #define mem0(a) memset(a,0,sizeof(a)) #define memmx(a) memset(a,0x3f3f,sizeof(a)) #define memmn(a) memset(a,-0x3f3f,sizeof(a)) #define debug #define db double #define yyes cout<<"YES"<<endl; #define nno cout<<"NO"<<endl; using namespace std; typedef vector<int> vei; typedef vector<pii> vep; typedef map<int,int> mpii; typedef map<char,int> mpci; typedef map<string,int> mpsi; typedef deque<int> deqi; typedef deque<char> deqc; typedef priority_queue<int> mxpq; typedef priority_queue<int,vector<int>,greater<int> > mnpq; typedef priority_queue<pii> mxpqii; typedef priority_queue<pii,vector<pii>,greater<pii> > mnpqii; const int maxn=500005; const int inf=0x3f3f3f3f3f3f3f3f; const int MOD=100000007; const db eps=1e-10; int qpow(int a,int b){int tmp=a%MOD,ans=1;while(b){if(b&1){ans*=tmp,ans%=MOD;}tmp*=tmp,tmp%=MOD,b>>=1;}return ans;} int lowbit(int x){return x&-x;} int max(int a,int b){return a>b?a:b;} int min(int a,int b){return a<b?a:b;} int mmax(int a,int b,int c){return max(a,max(b,c));} int mmin(int a,int b,int c){return min(a,min(b,c));} void mod(int &a){a+=MOD;a%=MOD;} bool chk(int now){} int half(int l,int r){while(l<=r){int m=(l+r)/2;if(chk(m))r=m-1;else l=m+1;}return l;} int ll(int p){return p<<1;} int rr(int p){return p<<1|1;} int mm(int l,int r){return (l+r)/2;} int lg(int x){if(x==0) return 1;return (int)log2(x)+1;} bool smleql(db a,db b){if(a<b||fabs(a-b)<=eps)return true;return false;} db len(db a,db b,db c,db d){return sqrt((a-c)*(a-c)+(b-d)*(b-d));} bool isp(int x){if(x==1)return false;if(x==2)return true;for(int i=2;i*i<=x;++i)if(x%i==0)return false;return true;} int t,sum[maxn]; char s[maxn]; signed main(){ ios_base::sync_with_stdio(0); cin.tie(0),cout.tie(0); sc(t); while(t--){ scanf("%s",s+1); int l=strlen(s+1); int ans=0; sum[0]=0; re(i,1,l) sum[i]=sum[i-1]+(s[i]=='0'); re(len,1,18){ for(int i=1;i+len-1<=l;++i){ if(s[i]=='1'){ int cal=0,base=1; rre(j,i+len-1,i){ if(s[j]=='1') cal+=base; base*=2; } if(cal>=len&&i-1>=(cal-len) &&sum[i-1]-sum[i-1-(cal-len)]>=(cal-len)) ++ans; } } } cout<<ans<<endl; } return 0; }
D、给一张有向图，每条边染色，要使得每个环都不能全部为同一种颜色，输出最少需要的颜色和每条边的颜色
先猜个结论，颜色要么一种要么两种，没环全涂1，否则把最后一条环上的边改成2
v[x]等于0代表未访问，等于1代表已访问但不在栈中，等于2代表已访问并且处于栈中
点数比较小，改颜色的操作我直接上二维数组存住
代码：
#include <bits/stdc++.h> #define int long long #define sc(a) scanf("%lld",&a) #define scc(a,b) scanf("%lld %lld",&a,&b) #define sccc(a,b,c) scanf("%lld %lld %lld",&a,&b,&c) #define schar(a) scanf("%c",&a) #define pr(a) printf("%lld",a) #define fo(i,a,b) for(int i=a;i<b;++i) #define re(i,a,b) for(int i=a;i<=b;++i) #define rfo(i,a,b) for(int i=a;i>b;--i) #define rre(i,a,b) for(int i=a;i>=b;--i) #define prn() printf("\n") #define prs() printf(" ") #define mkp make_pair #define pii pair<int,int> #define pub(a) push_back(a) #define pob() pop_back() #define puf(a) push_front(a) #define pof() pop_front() #define fst first #define snd second #define frt front() #define bak back() #define mem0(a) memset(a,0,sizeof(a)) #define memmx(a) memset(a,0x3f3f,sizeof(a)) #define memmn(a) memset(a,-0x3f3f,sizeof(a)) #define debug #define db double #define yyes cout<<"YES"<<endl; #define nno cout<<"NO"<<endl; using namespace std; typedef vector<int> vei; typedef vector<pii> vep; typedef map<int,int> mpii; typedef map<char,int> mpci; typedef map<string,int> mpsi; typedef deque<int> deqi; typedef deque<char> deqc; typedef priority_queue<int> mxpq; typedef priority_queue<int,vector<int>,greater<int> > mnpq; typedef priority_queue<pii> mxpqii; typedef priority_queue<pii,vector<pii>,greater<pii> > mnpqii; const int maxn=500005; const int inf=0x3f3f3f3f3f3f3f3f; const int MOD=100000007; const db eps=1e-10; int qpow(int a,int b){int tmp=a%MOD,ans=1;while(b){if(b&1){ans*=tmp,ans%=MOD;}tmp*=tmp,tmp%=MOD,b>>=1;}return ans;} int lowbit(int x){return x&-x;} int max(int a,int b){return a>b?a:b;} int min(int a,int b){return a<b?a:b;} int mmax(int a,int b,int c){return max(a,max(b,c));} int mmin(int a,int b,int c){return min(a,min(b,c));} void mod(int &a){a+=MOD;a%=MOD;} bool chk(int now){} int half(int l,int r){while(l<=r){int m=(l+r)/2;if(chk(m))r=m-1;else l=m+1;}return l;} int ll(int p){return p<<1;} int rr(int p){return p<<1|1;} int mm(int l,int r){return (l+r)/2;} int lg(int x){if(x==0) return 1;return (int)log2(x)+1;} bool smleql(db a,db b){if(a<b||fabs(a-b)<=eps)return true;return false;} db len(db a,db b,db c,db d){return sqrt((a-c)*(a-c)+(b-d)*(b-d));} bool isp(int x){if(x==1)return false;if(x==2)return true;for(int i=2;i*i<=x;++i)if(x%i==0)return false;return true;} int n,m,k=1; int x[maxn],y[maxn]; int p[5005][5005]; int ans[maxn]; int v[maxn]; void dfs(int x,int f){ if(v[x]==2){ p[f][x]=2; return; } v[x]=2; re(i,1,n){ if(p[x][i]!=0){ int y=i; if(v[y]==0) dfs(y,x); else if(v[y]==2) k=2,p[x][y]=2; } } v[x]=1; } signed main(){ ios_base::sync_with_stdio(0); cin.tie(0),cout.tie(0); cin>>n>>m; re(i,1,m) cin>>x[i]>>y[i],p[x[i]][y[i]]=1; re(i,1,n) if(v[i]==0) dfs(i,0); cout<<k<<endl; re(i,1,m) cout<<p[x[i]][y[i]]<<' '; return 0; } /* 3 6 1 2 2 1 1 3 3 1 2 3 3 2 */