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Music Notes(前缀和+二分)

發(fā)布時間:2024/9/3 编程问答 32 豆豆
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鏈接:https://ac.nowcoder.com/acm/contest/1077/I
來源:牛客網(wǎng)

Music Notes
時間限制:C/C++ 1秒,其他語言2秒
空間限制:C/C++ 32768K,其他語言65536K
64bit IO Format: %lld
題目描述
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).
Consider this song with three notes of durations 2, 1, and 3 beats:
Beat: 0 1 2 3 4 5 6 …
|----|----|----|----|----|----|— …
1111111111 : :
22222: :
333333333333333:

Here is a set of five queries along with the resulting answer:
Query Note

2 2
3 3
4 3
0 1
1 1

輸入描述:

  • 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
    輸出描述:
  • Lines 1…Q: Line i of the output contains the result of query i as a single integer.
    示例1
    輸入
    復(fù)制
3 521323401

輸出
復(fù)制

2 3 3 1 1

/*
這題時這場比賽的弱化版,下一題是強化版,這份代碼均適用。
x先余前綴最大和,
然后二分第一個大于等于,
后面特判一下,如果等于的話,res+1即可。

*/

#include <iostream>using namespace std; const int MAXN = 5e4+5; int s[MAXN]; int main() {int n,q;cin>>n>>q;int x;s[0] = 0;for(int i = 1; i <= n; i++){cin>>x;s[i] = s[i-1] + x;}while(q--){cin>>x;x %= s[n];int l = 1, r = n,res = 0;while(l <= r){int mid = (l + r) >> 1;if(x <= s[mid]){res = mid;r = mid - 1;}else l = mid + 1;}int ans = 0;if(x < s[res])ans = res;elseans = res + 1;cout<<ans<<endl;}return 0; }

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