C. Ayoub's function

考虑全为0的串数。我们有m个1，相当于有m+1的空隙可以用来放0。那么显然我们把这n-m个0匀着放到这m+1个空隙，可以使得每个0的部分尽可能短，进而使得全为0的子串尽可能少。

所以每个空隙放b=(n-m)/(m+1)。因为不能整除，所有有(n-m) % (m+1)个空隙实际上还要再多一个0。

如果每个空隙都放a = (n-m)/(m+1)个，那么显然每个空隙产生  a * (a + 1) / 2，一共产生(m + 1) * a * (a + 1) / 2 个全为0的子串。但是有b个空隙实际上还有一个0，所以还要在多产生b * (a+1)个串。

#include <iostream>
#include <vector>
#include <algorithm>
#include <string>
#include <set>
#include <queue>
#include <map>
#include <sstream>
#include <cstdio>
#include <cstring>
#include <numeric>
#include <cmath>
#include <iomanip>
#include <deque>
#include <bitset>
//#include <unordered_set>
//#include <unordered_map>
//#include <bits/stdc++.h>
//#include <xfunctional>
#define ll              long long
#define PII             pair<int, int>
#define rep(i,a,b)      for(int  i=a;i<=b;i++)
#define dec(i,a,b)      for(int  i=a;i>=b;i--)
#define pb              push_back
#define mk              make_pair
using namespace std;
int dir1[6][2] = { { 0,1 } ,{ 0,-1 },{ 1,0 },{ -1,0 },{ 1,1 },{ -1,1 } };
int dir2[6][2] = { { 0,1 } ,{ 0,-1 },{ 1,0 },{ -1,0 },{ 1,-1 },{ -1,-1 } };
const long long INF = 0x7f7f7f7f7f7f7f7f;
const int inf = 0x3f3f3f3f;
const double pi = 3.14159265358979;
const int mod = 100007;
const int N = 1005;
//if(x<0 || x>=r || y<0 || y>=c)

inline ll read()
{
    ll x = 0; bool f = true; char c = getchar();
    while (c < '0' || c > '9') { if (c == '-') f = false; c = getchar(); }
    while (c >= '0' && c <= '9') x = (x << 1) + (x << 3) + (c ^ 48), c = getchar();
    return f ? x : -x;
}
ll gcd(ll m, ll n)
{
    return n == 0 ? m : gcd(n, m % n);
}

int main()
{
    int T;
    cin >> T;
    while (T--)
    {
        int n, m;
        cin >> n >> m;
        ll tot = (ll)n * (n + 1) / 2;
        ll a = (n - m) / (m + 1);
        ll b = (n - m) % (m + 1);
        printf("%I64d
", tot - a * (a + 1) / 2 * (m + 1) - (a + 1) * b);
    }
    return 0;
}