A 夹娃娃

题目地址：

https://ac.nowcoder.com/acm/contest/5881/A

基本思路：

这题比较简单，算是签到题。我们直接计算出前缀和然后每次的查询区间和就行了。

参考代码：

#pragma GCC optimize(2)
#pragma GCC optimize(3)
#include <bits/stdc++.h>
using namespace std;
#define IO std::ios::sync_with_stdio(false)
#define ll long long
#define rep(i, l, r) for (int i = l; i <= r; i++)
#define per(i, l, r) for (int i = l; i >= r; i--)
#define mset(s, _) memset(s, _, sizeof(s))
#define pb push_back
#define pii pair <int, int>
#define mp(a, b) make_pair(a, b)
#define INF (int)1e18

inline int read() {
  int x = 0, neg = 1; char op = getchar();
  while (!isdigit(op)) { if (op == '-') neg = -1; op = getchar(); }
  while (isdigit(op)) { x = 10 * x + op - '0'; op = getchar(); }
  return neg * x;
}
inline void print(int x) {
  if (x < 0) { putchar('-'); x = -x; }
  if (x >= 10) print(x / 10);
  putchar(x % 10 + '0');
}

const int maxn = 1e5 + 10;
int n,k,sum[maxn];
signed main() {
  IO;
  n = read(), k = read();
  for (int i = 1; i <= n; ++i) {
    int x = read();
    sum[i] = sum[i - 1] + x;
  }
  while (k--) {
    int l = read(), r = read();
    int res = sum[r] - sum[l - 1];
    print(res);
    puts("");
  }
  return 0;
}