Codeforces 1301 C . Ayoub's function（容斥+贪心）

Description：

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 $s_{l},s_{l+1},…,s_{r}$ 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 $t (1≤t≤10^5)$ — the number of test cases. The description of the test cases follows.

The only line for each test case contains two integers $n, m (1≤n≤10^9, 0≤m≤n)$ — 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 $"1"$.

Example

input

5
3 1
3 2
3 3
4 0
5 2

output

4
5
6
0
12

Note

In the first test case, there exists only $3$ strings of length $3$, which has exactly 1 symbol, equal to $"1"$. These strings are: $s_{1}="100", s_{2}="010", s_{3}="001".$ The values of f for them are: $f(s_{1})=3,f(s_{2})=4,f(s_{3})=3$, so the maximum value is $4$ and the answer is $4$.

In the second test case, the string s with the maximum value is $"101".$

In the third test case, the string $s$ with the maximum value is $"111".$

In the fourth test case, the only string s of length 4, which has exactly $0$ symbols, equal to $"1"$ is $"0000"$ and the value of f for that string is $0$, so the answer is $0.$

In the fifth test case, the string s with the maximum value is $"01010"$ and it is described as an example in the problem statement.

题意：

给吹一个长度为 $n$ 的 $01$ 串，里面有 $m$ 个 $1$ ，可以任意选取区间，如果这个区间内至少含有一个 $1$ ，则得一分，问怎么排列可以使得得分最高，输出最高得分。

我们知道如果都是1的话我们可以选取 $\frac{n*(n+1)}{2}$ 对,那么如果有连续的 $0$ 那么假设 $0$ 的长度为 $l$ 那不能选取的对数就是 $\frac{l*(l+1)}{2}$ 。将这些 $0$ 划分为 $m + 1$ 个组，以便使每个组的 $\frac{l*(l+1)}{2}$ 之和最小。

最优的方法是将它们分成相等的组或尽可能相等。

每组的长度就是 $\frac{n-m}{m + 1}$ ,令 $k=\frac{n-m}{m + 1}$

这些不能选取的区间总数就是 $(m + 1) *\frac{k * (k + 1)}{2}$

如果不能整除划分区间，就把多余的 $0$ 分到每个区间内，这样加上这些 $0$ 对每个区间 不能选取的对数就变成了 $k+1$ 。全部的就是 $(z \% (m + 1)) * (k + 1)$

用全部可选的减去不能选取的就是最后答案。

AC代码：

#include <cstdio>
#include <vector>
#include <queue>
#include <cstring>
#include <cmath>
#include <map>
#include <set>
#include <string>
#include <iostream>
#include <algorithm>
#include <iomanip>
#include <stack>
#include <queue>
using namespace std;
#define sd(n) scanf("%d", &n)
#define sdd(n, m) scanf("%d%d", &n, &m)
#define sddd(n, m, k) scanf("%d%d%d", &n, &m, &k)
#define pd(n) printf("%d\n", n)
#define pc(n) printf("%c", n)
#define pdd(n, m) printf("%d %d\n", n, m)
#define pld(n) printf("%lld\n", n)
#define pldd(n, m) printf("%lld %lld\n", n, m)
#define sld(n) scanf("%lld", &n)
#define sldd(n, m) scanf("%lld%lld", &n, &m)
#define slddd(n, m, k) scanf("%lld%lld%lld", &n, &m, &k)
#define sf(n) scanf("%lf", &n)
#define sc(n) scanf("%c", &n)
#define sff(n, m) scanf("%lf%lf", &n, &m)
#define sfff(n, m, k) scanf("%lf%lf%lf", &n, &m, &k)
#define ss(str) scanf("%s", str)
#define rep(i, a, n) for (int i = a; i <= n; i++)
#define per(i, a, n) for (int i = n; i >= a; i--)
#define mem(a, n) memset(a, n, sizeof(a))
#define debug(x) cout << #x << ": " << x << endl
#define pb push_back
#define all(x) (x).begin(), (x).end()
#define fi first
#define se second
#define mod(x) ((x) % MOD)
#define gcd(a, b) __gcd(a, b)
#define lowbit(x) (x & -x)
typedef pair<int, int> PII;
typedef long long ll;
typedef unsigned long long ull;
typedef long double ld;
const int MOD = 1e9 + 7;
const double eps = 1e-9;
const ll INF = 0x3f3f3f3f3f3f3f3fll;
const int inf = 0x3f3f3f3f;
inline int read()
{
	int ret = 0, sgn = 1;
	char ch = getchar();
	while (ch < '0' || ch > '9')
	{
		if (ch == '-')
			sgn = -1;
		ch = getchar();
	}
	while (ch >= '0' && ch <= '9')
	{
		ret = ret * 10 + ch - '0';
		ch = getchar();
	}
	return ret * sgn;
}
inline void Out(int a) //Êä³öÍâ¹Ò
{
	if (a > 9)
		Out(a / 10);
	putchar(a % 10 + '0');
}

ll gcd(ll a, ll b)
{
	return b == 0 ? a : gcd(b, a % b);
}

ll lcm(ll a, ll b)
{
	return a * b / gcd(a, b);
}
///快速幂m^k%mod
ll qpow(ll a, ll b, ll mod)
{
	if (a >= mod)
		a = a % mod + mod;
	ll ans = 1;
	while (b)
	{
		if (b & 1)
		{
			ans = ans * a;
			if (ans >= mod)
				ans = ans % mod + mod;
		}
		a *= a;
		if (a >= mod)
			a = a % mod + mod;
		b >>= 1;
	}
	return ans;
}

// 快速幂求逆元
int Fermat(int a, int p) //费马求a关于b的逆元
{
	return qpow(a, p - 2, p);
}

///扩展欧几里得
int exgcd(int a, int b, int &x, int &y)
{
	if (b == 0)
	{
		x = 1;
		y = 0;
		return a;
	}
	int g = exgcd(b, a % b, x, y);
	int t = x;
	x = y;
	y = t - a / b * y;
	return g;
}

///使用ecgcd求a的逆元x
int mod_reverse(int a, int p)
{
	int d, x, y;
	d = exgcd(a, p, x, y);
	if (d == 1)
		return (x % p + p) % p;
	else
		return -1;
}

///中国剩余定理模板0
ll china(int a[], int b[], int n) //a[]为除数，b[]为余数
{
	int M = 1, y, x = 0;
	for (int i = 0; i < n; ++i) //算出它们累乘的结果
		M *= a[i];
	for (int i = 0; i < n; ++i)
	{
		int w = M / a[i];
		int tx = 0;
		int t = exgcd(w, a[i], tx, y); //计算逆元
		x = (x + w * (b[i] / t) * x) % M;
	}
	return (x + M) % M;
}

ll n, m;
int t;
ll ans, res, tmp, cnt;

int main()
{
	sd(t);
	while (t--)
	{
		sldd(n, m);
		ans = n * (n + 1) / 2;
		ll z = n - m;
		ll k = z / (m + 1);
		ans -= (m + 1) *k * (k + 1) / 2;                     
		ans -= (z % (m + 1)) * (k + 1);
		pld(ans);
	}
	return 0;
}