【牛客小白月赛6】F 发电

题目地址：https://www.nowcoder.com/acm/contest/136/F

树状数组、快速幂、逆元的模板运用；

#include<iostream>
#include<cstdio>
using namespace std;
 
#define LL long long
#define lowbit(x) x&-x
const int N = 1e6+5;
const LL MOD = 1e9+7;
int n, m;
LL sum[N];
 
void read(int &x) {
    int f = 1; x = 0;
    char ch = getchar();
    
    while (ch < '0' || ch > '9') {if (ch == '-') f = -1; ch = getchar();}
    while (ch >= '0' && ch <= '9') {x = x * 10 + ch - '0'; ch = getchar();}
    x *= f;
}
 
LL quick_pow(LL a, LL k)
{
    LL tmp=1;
    while(k) {
        if(k&1)
            tmp = tmp*a%MOD;
        a = a*a%MOD;
        k>>=1;
    }
    return tmp;
}
 
LL find(LL b)
{
    LL sum1=1;
    for(int i=b; i; i-=lowbit(i))
        sum1 = sum1*sum[i]%MOD;
    return sum1;
}
 
int main()
{
    read(n); read(m);
    for(int i=1; i<=n; i++) sum[i]=1;
    
    for(int x,y,z,t=0; t<m; t++) {
        read(x); read(y); read(z);
         
        if(x == 1) {
            for(int i=y; i<=n; i+=lowbit(i))
                sum[i] = sum[i]*z%MOD;
        }
        else if(x == 2) {
            LL inv = quick_pow(z, MOD-2);
            for(int i=y; i<=n; i+=lowbit(i))
                sum[i] = sum[i]*inv%MOD;
        }
        else if(x == 3) {
            LL ans = find(z)*quick_pow(find(y-1), MOD-2);
            ans = (ans%MOD + MOD)%MOD;
            printf("%lld
", ans);
        }
    }
    
    return 0;
}