Codeforces 1062C Banh-mi题解

思路：
$$每次吃x_i最大的食物$$

$$假设初始区间内有a个1，b个0，那么先吃完所有初始美味为1的食物，得到喜悦程度为2^a-1$$

人后剩下的就是初始美味为0的食物，现在美味度为
$$2^a-1$$
所以吃完这些喜悦度就是
$$(2^a-1)(2^b-1)$$
然后通过前缀和求出任意区间01数目，利用快速幂求下式即可
$$(2^a-1)2^b$$

#include<cstdio>
using namespace std;
typedef long long ll;
const ll mod = 1e9 + 7;
const int maxn = 1e5 + 10;
int n, q;
char s[maxn];
int c[2][maxn];
ll fpow(ll a, ll n)
{
   
	ll res = 1, base = a % mod;
	while (n)
	{
   
		if (n & 1) res *= base, res %= mod;
		base *= base, base %= mod;
		n >>= 1;
	}
	return res % mod;
}
int main(void){
   
	scanf("%d %d", &n, &q);
	scanf("%s", s + 1);
	c[0][0] = c[1][0] = 0;
	for (int i = 1, x; i <= n; i++)
	{
   
		x = s[i] - '0';
		c[x][i] = c[x][i - 1] + 1;
		c[x ^ 1][i] = c[x ^ 1][i - 1];
	}
	for (int i = 1, l, r; i <= q; i++)
	{
   
		scanf("%d %d", &l, &r);
		ll a = c[1][r] - c[1][l - 1], b = c[0][r] - c[0][l - 1];
		printf("%lld\n", (fpow(2, a + b) - fpow(2, b) + mod) % mod);
	}
	return 0;
}