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; }