[USACO 2009 Dec S]Music Notes

⭐简单说两句⭐

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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).

在这里插入图片描述

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;
}

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结语

谢谢你的阅读，由于作者水平有限，难免有不足之处，若读者发现问题，还请批评，在留言区留言或者私信告知，我一定会尽快修改的。若各位大佬有什么好的解法，或者有意义的解法都可以在评论区展示额，万分谢谢。
写作不易，望各位老板点点赞，加个关注!😘😘😘

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作者：后端小知识

CSDN个人主页：后端小知识

🔎GZH：后端小知识

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