Poj3468 A Simple Problem with Integers （分块）

题面

Poj

题解

区间求和$+$区间修改板子，这里用分块写的

#include <cmath>
#include <cstdio>
#include <cstring>
#include <algorithm>
using std::min; using std::max;
using std::swap; using std::sort;
typedef long long ll;
#define int ll

const int N = 1e5 + 10 , SN = 340;
int n, siz, q, bel[N], val[N];
int sum[SN], add[SN], L[SN], R[SN];

template<typename T>
void read(T &x) {
	int flag = 1; x = 0; char ch = getchar();
	while(ch < '0' || ch > '9') { if(ch == '-') flag = -flag; ch = getchar(); }
	while(ch >= '0' && ch <= '9') x = x * 10 + ch - '0', ch = getchar(); x *= flag;
}

void modify (int l, int r, int c) {
	int fl = bel[l], fr = bel[r];
	if(fl == fr) {
		for(int i = l; i <= r; ++i)
			val[i] += c, sum[fl] += c;
	} else {
		for(int i = l; i <= R[fl]; ++i)
			val[i] += c, sum[fl] += c;
		for(int i = fl + 1; i < fr; ++i) add[i] += c;
		for(int i = L[fr]; i <= r; ++i)
			val[i] += c, sum[fr] += c;
	}
}

int query(int l, int r) {
	int fl = bel[l], fr = bel[r], ret = 0;
	if(fl == fr) {
		for(int i = l; i <= r; ++i)
			ret += val[i] + add[fl];
	} else {
		for(int i = l; i <= R[fl]; ++i)
			ret += val[i] + add[fl];
		for(int i = fl + 1; i < fr; ++i) ret += sum[i] + add[i] * (R[i] - L[i] + 1);
		for(int i = L[fr]; i <= r; ++i)
			ret += val[i] + add[fr];
	} return ret;
}

signed main () {
	read(n), read(q), siz = sqrt(n);
	for(int i = 1; i <= n; ++i)
		read(val[i]), bel[i] = (i - 1) / siz + 1, sum[bel[i]] += val[i];
	for(int i = 1; i <= bel[n]; ++i)
		L[i] = R[i - 1] + 1, R[i] = i * siz;
	R[bel[n]] = n; int l, r, k;
	while(q--) {
		char opt; scanf("\n%c", &opt);
		read(l), read(r);
		if(opt == 'Q') printf("%lld\n", query(l, r));
		else read(k), modify(l, r, k);
	}
	return 0;
}