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.

题意
给定一个长度为n的只包含0和1字符串，其中有m个1，要求在字符串人取两个数l,r，l<=r，并且在[l,r]的范围内至少有一个1。
问最多有几种取l和r的取法。

思路
以内对字符串任意排列，我们很容易得知，所有的1分散开得到的一定最多。
假设有7个，2个0。
我们有两种排法：0101000或者0010100
我们知道
$C_3^2>2*C^2_2$
所以根据贪心规则，将所有的0平分去插1的空是最多的。
用一点容斥的思想。
$ans=C_n^2-sigma(每个空0的组合数)+num(1)$

AC代码

#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;
}