题意略。

思路：这题考察的是二分搜索。

#include
     
      #define maxn 200005using namespace std;typedef long long LL;LL n,q;LL ai[maxn],ki[maxn];int main(){    scanf("%lld%lld",&n,&q);    for(int i = 0;i < n;++i) scanf("%lld",&ai[i]);    for(int i = 1;i <= q;++i) scanf("%lld",&ki[i]);        for(int i = 1;i < n;++i) ai[i] += ai[i - 1];        LL sum = 0;    for(int i = 1;i <= q;++i){        sum += ki[i];        int die = upper_bound(ai,ai + n,sum) - ai;        int live = n - die;        if(live == 0){            live = n;            sum = 0;        }        printf("%d\n",live);    }    return 0;}