[USACO 2009 Dec S]Music Notes

题目： [USACO 2009 Dec S]Music Notes ，哈哈，我们今天来看一道有二分思想的题嘛，这是选自USACO上的一道题，好了，我们一起来看看题意吧：

题目描述是复制的，可能有部分显示不对，我就把题目链接放下面！

题目链接： [USACO 2009 Dec S]Music Notes

题目描述

FJ is going to teach his cows how to play a song. The song consists of N (1 <= N <= 50,000) notes, and the i-th note lasts for Bi (1 <= Bi <= 10,000) beats (thus no song is longer than 500,000,000 beats). The cows will begin playing the song at time 0; thus, they will play note 1 from time 0 through just before time B1, note 2 from time B1 through just before time B1 + B2, etc. However, recently the cows have lost interest in the song, as they feel that it is too long and boring. Thus, to make sure his cows are paying attention, he asks them Q (1 <= Q <= 50,000) questions of the form, "In the interval from time T through just before time T+1, which note should you be playing?" The cows need your help to answer these questions which are supplied as Ti (0 <= Ti <= end_of_song).

输入描述

• Line 1: Two space-separated integers: N and Q
• Lines 2…N+1: Line i+1 contains the single integer: Bi
• Lines N+2…N+Q+1: Line N+i+1 contains a single integer: Ti

输出描述

• Lines 1…Q: Line i of the output contains the result of query i as a single integer.

示例1

输入

3 5

2

1

3

2

3

4

0

1

输出

2

3

3

1

1

思路:

采用前缀和与二分的思想，很快就解决了

我们来看看成功AC的代码吧：

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
int n,q;
int a[1000010];
int sum[1000010];//存放前缀和的
int main(){
    ios::sync_with_stdio(false);
    cin>>n>>q;
    for(int i=1;i<=n;i++){
        cin>>a[i];
        sum[i]=sum[i-1]+a[i];
    }
    while(q--){
        int x;  cin>>x;
        int ans=upper_bound(sum+1,sum+n+1,x)-sum;//这里减掉起始位置就是元素的下标了
        cout<<ans<<"\n";
    }
    return 0;
}