(Jizhong) 1592. [Training] GDKOI musical beats (mnotes) [] + analog-half

(File IO): input: mnotes.in output: mnotes.out
time limit: 1000 ms space constraints: 131072 KB specific restrictions
Goto ProblemSet

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Title Description
$FJ$ ready to teach his cows playing a song, the song $N(1<=N<=50,000)$ species syllables, numbered $1$ to $N$ , and must follow from $1$ to $N$ is the order of play for the first $i$ kind of sustained syllables $B_i(1<=B_i<=10,000)$ beats, beats from $0$ starts counting, so beat from $0$ to beat $B_1-1$ play of the first $1$ Zhong syllable, from $B_1$To $B_1+B_2-1$ play of the first $2$ Zhong syllable, and so on.
Cows are not interested in playing for recently, and they feel too boring. So in order to keep the focus on the cows, $FJ$ proposed $Q(1<=Q<=50,000)$ questions, format questions is "first $T$ times beats playing what kind of syllables, "
each corresponding to a problem $T_i (0 <= T_i <= total number of beats -1)$ Please help to resolve the cows.

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Input
first line of input two spaces separated by an integer $N$ and $Q$
first $2$ to $N+1$ lines contains an integer $B_i$
The first $N+2-N+Q+1$ lines contains an integer $T_i$

Output
Output has $Q$ lines, each line of output corresponding to the answer to the question.

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Sample input
. 3. 5
2
. 1
. 3
2
. 3
. 4
0
. 1

Sample Output
2
. 3
. 3
. 1
. 1

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Data range limit

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Solving ideas
bipartite + analog

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Code

#include<iostream>
#include<cstring>
#include<string>
#include<cstdio>
#include<algorithm>
#include<iomanip>
#include<cmath>
using namespace std;
int n,q,a[50010],x,l,r,mid;
bool flag;
int main(){
    freopen("mnotes.in","r",stdin);
    freopen("mnotes.out","w",stdout);
    scanf("%d%d",&n,&q);
    for(int i=1;i<=n;i++)
    {
        scanf("%d",&a[i]);
        a[i]+=a[i-1];
    }
    for(int i=1;i<=q;i++)
    {
        scanf("%d",&x);
        ++x;
        l=1;r=n;
        flag=0;
        while(l<r)
        {
            mid=(l+r)/2;
            if(a[mid]>x)
                r=mid;
            if(a[mid]<x)
                l=mid+1;
            if(a[mid]==x)
            {
                printf("%d\n",mid);
                flag=1;
                break;
                
            }
        }
        if(!flag)
            printf("%d\n",l);
    }
}