Codeforces Round #619 (Ayoub's function)

题目：
Ayoub thinks that he is a very smart person, so he created a function f(s), where s is a binary string (a string which contains only symbols “0” and “1”). The function f(s) is equal to the number of substrings in the string s that contains at least one symbol, that is equal to “1”.
More formally, f(s) is equal to the number of pairs of integers (l,r), such that 1≤l≤r≤|s| (where |s| is equal to the length of string s), such that at least one of the symbols sl,sl+1,…,sr is equal to “1”.
For example, if s=“01010” then f(s)=12, because there are 12 such pairs (l,r): (1,2),(1,3),(1,4),(1,5),(2,2),(2,3),(2,4),(2,5),(3,4),(3,5),(4,4),(4,5).
Ayoub also thinks that he is smarter than Mahmoud so he gave him two integers n and m and asked him this problem. For all binary strings s of length n which contains exactly m symbols equal to “1”, find the maximum value of f(s).
Mahmoud couldn’t solve the problem so he asked you for help. Can you help him?

Input
The input consists of multiple test cases. The first line contains a single integer t (1≤t≤105) — the number of test cases. The description of the test cases follows.
The only line for each test case contains two integers n, m (1≤n≤109, 0≤m≤n) — the length of the string and the number of symbols equal to “1” in it.

Output
For every test case print one integer number — the maximum value of f(s) over all strings s of length n, which has exactly m symbols, equal to “1”.

样例：

输入：
5
3 1
3 2
3 3
4 0
5 2
输出：
4
5
6
0
12

思路：
推出公式即可求解：
(a * (a + 1) / 2) - (c * (c + 1) / 2) * d - ((c + 1) * (c + 2) / 2) * (b + 1 - d)；
其中c = (a - b) / (b + 1); d = (b + 1 - (a - b) % (b + 1));

AC代码：

#include<stdio.h>
using namespace std;
#define ll long long
const int maxn = 1e5 + 5;
const int inf = 0x3f3f3f3f;
int a[maxn], b[maxn];
void solve(ll a, ll b) {
	ll c = (a - b) / (b + 1);
	ll d = (b + 1 - (a - b) % (b + 1));
	printf("%lld\n", (a * (a + 1) / 2) - (c * (c + 1) / 2) * d - ((c + 1) * (c + 2) / 2) * (b + 1 - d));
}
int main()
{
	int t; scanf("%d", &t);
	for (int i = 0; i < t; i++) {
		ll a, b;
		//防止乘爆，用ll
		scanf("%lld%lld", &a, &b);
		solve(a, b);
	}
	return 0;
}