前缀和+STL

前缀和应该挺简单不用我说吧，C++中一个STL叫upper_bound()查找在一个左闭右开的区间中第一个大于x的迭代器位置，记住不带等号！！带等号的是lower_bound()！！！迭代器减掉起始下标就是距离了。

#pragma GCC target("avx,sse2,sse3,sse4,popcnt")
#pragma GCC optimize("O2,O3,Ofast,inline,unroll-all-loops,-ffast-math")
#include <bits/stdc++.h>
using namespace std;
#define js ios::sync_with_stdio(false);cin.tie(0); cout.tie(0)
#define all(vv) (vv).begin(), (vv).end()
#define endl "\n"
typedef long long ll; typedef unsigned long long ull; typedef long double ld;
const ll MOD = 1e9 + 7;
inline ll read() { ll s = 0, w = 1; char ch = getchar(); for (; !isdigit(ch); ch = getchar()) if (ch == '-') w = -1; for (; isdigit(ch); ch = getchar())    s = (s << 1) + (s << 3) + (ch ^ 48); return s * w; }
inline void write(ll x) { if (!x) { putchar('0'); return; } char F[40]; ll tmp = x > 0 ? x : -x; if (x < 0)putchar('-');    int cnt = 0;    while (tmp > 0) { F[cnt++] = tmp % 10 + '0';        tmp /= 10; }    while (cnt > 0)putchar(F[--cnt]); }
inline ll gcd(ll x, ll y) { return y ? gcd(y, x % y) : x; }
ll qpow(ll a, ll b) { ll ans = 1;    while (b) { if (b & 1)    ans *= a;        b >>= 1;        a *= a; }    return ans; }    ll qpow(ll a, ll b, ll mod) { ll ans = 1; while (b) { if (b & 1)(ans *= a) %= mod; b >>= 1; (a *= a) %= mod; }return ans % mod; }
inline int lowbit(int x) { return x & (-x); }

const int N = 1e6 + 7;
int a[N];

int main() {
    int n = read(), m = read();
    for (int i = 1; i <= n; ++i)    a[i] = a[i - 1] + read();
    while (m--) {
        int x = read();
        int ans = upper_bound(a + 1, a + 1 + n, x) - a;
        write(ans), putchar(10);
    }
    return 0;
}