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;
}