C++(g++ 7.5.0) 解法, 执行用时: 51ms, 内存消耗: 1584K, 提交时间: 2022-10-20 17:25:19

#include <bits/stdc++.h>
using namespace std;
int t;
long long s,c,k,ans;
int main() {
    scanf("%d",&t);
    for (int o=1;o<=t;++o) {
        scanf("%lld%lld",&s,&c);
        if(c>s) {
            printf("%lld\n",c*s);
        }
        else {
            k=min(c/2,(s+1)/c);
            ans=((c-k)*c*k+s*(s+1))/2;
            k=(s-1)/(c+1)+1;
            ans=max(ans,c*k*s-k*(k-1)*c*(c+1)/2);
            printf("%lld\n",ans);
        }
    }
    return 0;
}

C++(clang++ 11.0.1) 解法, 执行用时: 50ms, 内存消耗: 1628K, 提交时间: 2022-09-26 22:40:29

#include <bits/stdc++.h>
using namespace std;
int T,s,c;
long long calc(int x){
    if (x<=0) return -1;
	int k=(x+c)/(c+1)-1;
	return (1ll*(s+x+1)*(s-x)+(2ll*x*c-1ll*k*(c+1)*c)*(k+1))/2;
}
void Main(){
	scanf("%d%d",&s,&c);
	int u=min(1ll*(c+1)*c/2,1ll*s);
	int x=u/c*c;
	printf("%lld\n",max(calc(x),max(calc(u),1ll*s*(s+1)/2)));
}
int main(){
	for (scanf("%d",&T);T--;Main());
	return 0;
}