Codeforces Round #619 (Div. 2) C. Ayoub's function(组合数学)
C. Ayoub's function
Ayoub thinks that he is a very smart person, so he created a function f(s), where ss is a binary string (a string which contains only symbols "0" and "1"). The function f(s)f(s) is equal to the number of substrings in the string ss 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||s| is equal to the length of string ss), such that at least one of the symbols sl,sl+1,…,srsl,sl+1,…,sr is equal to "1".
For example, if s=s="01010" then f(s)=12, because there are 12 such pairs (l,r)(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 nn and mm and asked him this problem. For all binary strings ss of length nn which contains exactly mm symbols equal to "1", find the maximum value of f(s)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.
OutputFor every test case print one integer number — the maximum value of f(s) over all strings ss of length nn, which has exactly mm symbols, equal to "1".
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 33 strings of length 33, which has exactly 11 symbol, equal to "1". These strings are: s1=s1="100", s2=s2="010", s3=s3="001". The values of ff for them are: f(s1)=3,f(s2)=4,f(s3)=3f(s1)=3,f(s2)=4,f(s3)=3, so the maximum value is 44 and the answer is 44.
In the second test case, the string ss with the maximum value is "101".
In the third test case, the string ss with the maximum value is "111".
In the fourth test case, the only string ss of length 44, which has exactly 00 symbols, equal to "1" is "0000" and the value of ff for that string is 00, so the answer is 00.
In the fifth test case, the string ss with the maximum value is "01010" and it is described as an example in the problem statement.链接：http://codeforces.com/contest/1301/problem/C
题意：二进制串长度为n、m个1，有几种组合使得每个串都包含1。
题解：组合数学，每两个1之间平分0的个数，该顺序才可算出最大值，然后用总量减去不可行量。
#include
using namespace std;
int main()
{
int t;
long long n,m;
scanf("%d",&t);
while(t--)
{
scanf("%I64d%I64d",&n,&m); //平均放 0
long long k=(n-m)/(m+1); //m+1个空段，每个空段平均K个0
long long l=(n-m)%(m+1); //还剩L个0
long long ans=n*(n+1)/2; //原始共ans个组合
long long len1=m+1-l; //len1个空段，每个段放K个
long long len2=l; //len2个空段，每个段放K+1个
ans=ans- k*(k+1)/2*len1 - (k+1)*(k+2)/2*len2;
cout<