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=100010;
const int INF=0x3f3f3f3f;
const int MOD=95041567,STA=8000010;
const LL LNF=1LL<<60;
const double EPS=1e-8;
const double OO=1e15;
const int dx[4]={-1,0,1,0};
const int dy[4]={0,1,0,-1};
const int day[13]={0,31,28,31,30,31,30,31,31,30,31,30,31};
//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)>>1;
        if(sum[mid]<tar)l=mid+1;
        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=1;i<=n;i++){
            scanf("%d",&a);
            sum[i]=sum[i-1]+a;
        }
        for(i=1;i<=m;i++)
            scanf("%d",&t[i]);
        w=ans=0;
        for(i=1;i<=m;i++){
            w+=t[i]-t[i-1]-1;
            if(w>=n){
                ans+=n-(w-(t[i]-t[i-1]-1));
                break;
            }
            ans=t[i];
            w=binary(1,w+1,sum[w]-k)-1;
        }
        if(w<n){
            ans+=n-w;
        }

        printf("%d
",ans);
    }
    return 0;
}