Color the ball

Time Limit: 9000/3000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 22899 Accepted Submission(s): 11132

Problem Description

N个气球排成一排，从左到右依次编号为1,2,3....N.每次给定2个整数a b(a <= b),lele便为骑上他的“小飞鸽"牌电动车从气球a开始到气球b依次给每个气球涂一次颜色。但是N次以后lele已经忘记了第I个气球已经涂过几次颜色了，你能帮他算出每个气球被涂过几次颜色吗？

Input

每个测试实例第一行为一个整数N,(N <= 100000).接下来的N行，每行包括2个整数a b(1 <= a <= b <= N)。
当N = 0，输入结束。

Output

每个测试实例输出一行，包括N个整数，第I个数代表第I个气球总共被涂色的次数。

Sample Input

3
1 1
2 2
3 3
3
1 1
1 2
1 3
0

Sample Output

1 1 1
3 2 1

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- 代码 -

#include<cstdio>
using namespace std;

typedef long long ll;
const int M = 2e5 + 5;

ll A[M], B[M], C[M];
int n, m, x, y, p, v;
int f;
char ch;

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

int lowbit(int x) { return x&(-x); }

void upt(int x, ll v) {
	int k = x;
	while (x <= n) {
		B[x] += v;
		C[x] += 1ll * v * k;
		x += lowbit(x);
	}
}

ll ask(int x) {
	ll ans = 0;
	int k = x;
	while (x > 0) {
		ans += (k + 1) * B[x] - C[x];
		x -= lowbit(x);
	}
	return ans;
}

int main() {
	read(n);
	for (int i = 1; i <= n; ++ i) {
		read(A[i]);
		upt(i, A[i] - A[i-1]);
	}
	read(m);
	for (int i = 1; i <= m; ++i) {
		read(p); read(x); read(y);
		if (p == 1) {
			read(v);
			upt(x, v); upt(y + 1, -v);
		}
		else printf("%lld
", ask(y) - ask(x - 1));
	}
	return 0;
}