URAL-1998 The old Padawan 二分

　　题目链接：http://acm.timus.ru/problem.aspx?space=1&num=1998

　　题意：有n个石头，每个石头有个重量，每个时间点你能让一个石头飞起来，但有m个时间点，你会分心，使得已经飞起来的石头会有重量之和大于k的石头掉下来，问你最终使的所有石头飞起来的时间。

　　维护一个前缀和，然后二分就可以了。

 //STATUS:C++_AC_93MS_1113KB
 #include <functional>
 #include <algorithm>
 #include <iostream>
 //#include <ext/rope>
 #include <fstream>
 #include <sstream>
 #include <iomanip>
 #include <numeric>
 #include <cstring>
 #include <cassert>
 #include <cstdio>
 #include <string>
 #include <vector>
 #include <bitset>
 #include <queue>
 #include <stack>
 #include <cmath>
 #include <ctime>
 #include <list>
 #include <set>
 #include <map>
 using namespace std;
 //#pragma comment(linker,"/STACK:102400000,102400000")
 //using namespace __gnu_cxx;
 //define
 #define pii pair<int,int>
 #define mem(a,b) memset(a,b,sizeof(a))
 #define lson l,mid,rt<<1
 #define rson mid+1,r,rt<<1|1
 #define PI acos(-1.0)
 //typedef
 typedef __int64 LL;
 typedef unsigned __int64 ULL;
 //const
 const int N=;
 const int INF=0x3f3f3f3f;
 const int MOD=,STA=;
 const LL LNF=1LL<<;
 const double EPS=1e-;
 const double OO=1e15;
 const int dx[]={-,,,};
 const int dy[]={,,,-};
 const int day[]={,,,,,,,,,,,,};
 //Daily Use ...
 inline int sign(double x){return (x>EPS)-(x<-EPS);}
 template<class T> T gcd(T a,T b){return b?gcd(b,a%b):a;}
 template<class T> T lcm(T a,T b){return a/gcd(a,b)*b;}
 template<class T> inline T lcm(T a,T b,T d){return a/d*b;}
 template<class T> inline T Min(T a,T b){return a<b?a:b;}
 template<class T> inline T Max(T a,T b){return a>b?a:b;}
 template<class T> inline T Min(T a,T b,T c){return min(min(a, b),c);}
 template<class T> inline T Max(T a,T b,T c){return max(max(a, b),c);}
 template<class T> inline T Min(T a,T b,T c,T d){return min(min(a, b),min(c,d));}
 template<class T> inline T Max(T a,T b,T c,T d){return max(max(a, b),max(c,d));}
 //End

 int t[N],sum[N];
 int n,m,k;

 int binary(int l,int r,LL tar)
 {
     int mid;
     while(l<r){
         mid=(l+r)>>;
         if(sum[mid]<tar)l=mid+;
         else r=mid;
     }
     return l;
 }

 int main()
 {
  //   freopen("in.txt","r",stdin);
     int i,j,a,w,ans;
     while(~scanf("%d%d%d",&n,&m,&k))
     {
         for(i=;i<=n;i++){
             scanf("%d",&a);
             sum[i]=sum[i-]+a;
         }
         for(i=;i<=m;i++)
             scanf("%d",&t[i]);
         w=ans=;
         for(i=;i<=m;i++){
             w+=t[i]-t[i-]-;
             if(w>=n){
                 ans+=n-(w-(t[i]-t[i-]-));
                 break;
             }
             ans=t[i];
             w=binary(,w+,sum[w]-k)-;
         }
         if(w<n){
             ans+=n-w;
         }

         printf("%d\n",ans);
     }
     return ;
 }