c/c++语言开发共享洛谷P3924 康娜的线段树(期望 前缀和)

题意

题目链接

sol

思路就是根据期望的线性性直接拿前缀和算贡献。。

这题输出的时候是不需要约分的qwq

如果你和我一样为了ac不追求效率的话直接#define int __int128就行了。。

代码十分清新

#include<bits/stdc++.h> #define int __int128 using namespace std; const int maxn = 1e6 + 10; inline int read() {     char c = getchar(); int x = 0, f = 1;     while(c < '0' || c > '9') {if(c == '-') f = -1; c = getchar();}     while(c >= '0' && c <= '9') x = x * 10 + c - '0', c = getchar();     return x * f; } void print(int x) {     if(x < 0) putchar('-'), x = -x;     if (x > 9) print(x / 10);     putchar('0' + x % 10); } int n, m, qwq, s[maxn], a[maxn], ans, lim; int get(int dep) {     return 1 << (lim - (dep - 1)); } void build(int l, int r, int dep, int sum) {     ans += (a[r] - a[l - 1]) * get(dep);     if(l == r) {s[l] = sum + get(dep); return ;}     int mid = l + r >> 1;     build(l, mid, dep + 1, sum + get(dep));      build(mid + 1, r,dep + 1, sum + get(dep)); } signed main() {     n = read(); m = read(); qwq = read();     for(int cur = 1; cur <= n; lim ++, cur <<= 1);     for(int i = 1; i <= n; i++) a[i] = read(), a[i] += a[i - 1];     build(1, n, 1, 0);     for(int i = 1; i <= n; i++) s[i] += s[i - 1];     while(m--) {         int l = read(), r = read(), v = read();         ans += ((s[r] - s[l - 1]) * v);         print(((ans * qwq) >> lim)); putchar('n');     }     return 0; }