Codeforces Round #619 (Div. 2)：C. Ayoub's function

Discription
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”.

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: s1=“100”, s2=“010”, s3=“001”. The values of f for them are: f(s1)=3,f(s2)=4,f(s3)=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.

Title intended
for a given length n 0 and 1 contain only string, wherein there are m 1, requiring two people to pick the string numbers l, r, l <= r , and the scope of the [l, r] of there are at least a 1.
Q. There are several borrowing up to take the l and r.

Thinking
within an arbitrary arrangement of strings, it is easy to learn, all one must spread out up to get.
Suppose there are 7, 0 2.
We have two styles: 0101000 or 0010100
us know
$C_3^2>2*C^2_2$
So greedy according to the rules, all to bisect 0 1 Interpolation is the largest.
With a little idea of inclusion and exclusion.
$ans = C_n ^ 2-sigma (per number of combinations of empty 0) + num (1)$

AC Code

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
#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 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 sff(n, m) scanf("%lf%lf", &n, &m)
#define sfff(n, m, k) scanf("%lf%lf%lf", &n, &m, &k)
#define pd(n) printf("%d\n", (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 rep(i, a, n) for(int i=a; i<=n; i++)
#define per(i, n, a) for(int i=n; i>=a; i--)
 
int T;
ll n,m,ans;
 
int main()
{
    scanf("%d",&T);
    while(T--)
    {
        sldd(n,m);
        if(m==0)
        {
            cout<<0<<endl;
            continue;
        }
        ans=n*(n-1)/2;
        ll tmp=n-m;
        ll xx=tmp/(m+1);
        //cout<<xx<<endl;
        ll tt=tmp%(m+1);
        //cout<<tt<<endl;
        ll cnt=(m+1-tt)*(xx-1)*xx/2+tt*(xx+1)*xx/2;
        ans-=cnt;
        ans+=m;
        cout<<ans<<endl;
    }
    return 0;
}