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

C. Ayoub’s function
time limit per test1 second
memory limit per test256 megabytes
inputstandard input
outputstandard output
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

inputCopy
5
3 1
3 2
3 3
4 0
5 2
outputCopy
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：给了一个长度为n的01串，其中有m个1，现在让你构造一个01串s，让他的f(s)最大，f(s)为字符1的个数≥1的子串个数

思路：首先一个字符串的子串个数自然就是n*(n+1)/2，我们要让f(s)最大，其实也就是等价 把n-m个0 平均分成(m+1)份 因为有m个1，每个1可以当作隔板一样隔开。
那么我们算出来num=(n-m)/(m+1) 这就是平均分的每一部分的0的个数 对于r=（n-m）%（m+1）
当然是在m+1份里 每个放1即可
那么答案就是 n*(n+1)/2-r*(num+1)(m+1-r)(num+2)/2-num(num+1)/2

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
#define me(a,x) memset(a,x,sizeof a)
#define pb(a) push_back(a)
#define pa pair<int,int>
#define fi first
#define se second
int main()
{

    int t;
    cin>>t;
    while(t--)
    {
        ll n,m;cin>>n>>m;
        ll ans=n*(n+1)/2;
        ll num=(n-m)/(m+1);
        ll r=(n-m)%(m+1);
        cout<<ans-r*(num+1)*(num+2)/2-(m+1-r)*num*(num+1)/2<<endl;
    }
    return 0;
}