【牛客小白月赛6】F 发电 - 树状数组&快速幂&逆元

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+;
 const LL MOD = 1e9+;
 int n, m;
 LL sum[N];

 void read(int &x) {
     int f = ; x = ;
     char ch = getchar();

     while (ch < '' || ch > '') {if (ch == '-') f = -; ch = getchar();}
     while (ch >= '' && ch <= '') {x = x *  + ch - ''; ch = getchar();}
     x *= f;
 }

 LL quick_pow(LL a, LL k)
 {
     LL tmp=;
     while(k) {
         if(k&)
             tmp = tmp*a%MOD;
         a = a*a%MOD;
         k>>=;
     }
     return tmp;
 }

 LL find(LL b)
 {
     LL sum1=;
     for(int i=b; i; i-=lowbit(i))
         sum1 = sum1*sum[i]%MOD;
     return sum1;
 }

 int main()
 {
     read(n); read(m);
     for(int i=; i<=n; i++) sum[i]=;

     for(int x,y,z,t=; t<m; t++) {
         read(x); read(y); read(z);

         if(x == ) {
             for(int i=y; i<=n; i+=lowbit(i))
                 sum[i] = sum[i]*z%MOD;
         }
         else if(x == ) {
             LL inv = quick_pow(z, MOD-);
             for(int i=y; i<=n; i+=lowbit(i))
                 sum[i] = sum[i]*inv%MOD;
         }
         else if(x == ) {
             LL ans = find(z)*quick_pow(find(y-), MOD-);
             ans = (ans%MOD + MOD)%MOD;
             printf("%lld\n", ans);
         }
     }

     return ;
 }