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Re:110ms水过In Reply To:110ms水过 Posted by:astoninfer at 2015-05-29 23:44:25 > 找规律的过程还是比较费神的,注意到k % i(i : 1→n)有分段等差数列即可,
> 另,数据在int边缘,不仅最终输出要用int64或longlong,中间过程只用int也会溢出
> #include <cstdio>
>
> using namespace std;
>
> int main(){
> __int64 k, n;
> while(~scanf("%I64d%I64d", &n, &k)){
> __int64 x = 0;
> for(int i = 1; i <= n; i++) x += k % i;
> printf("check : %I64d\n", x);
> __int64 ans = 0;
> if(n > k) { ans = (n - k) * k; n = k; }
> __int64 p;
> for(int i = 1; ; i++) if(k / i <= k / ( i + 1) + 2) { p = i; break; }
> if(n <= k / (p + 1)){
> for(int i = 1; i <= n; i++) ans += k % i;
> printf("%I64d\n", ans);
> continue;
> }
> for(int i = p; i >= 1; i--){
> int cnt = k / i - k / (i + 1);
> int t = k % (k / i) + (cnt - 1) * i;
> if(n < k / (i + 1) + 1) break;
> if(n < k / i) cnt = n - k / (i + 1);
> ans += cnt * (2 * t + (1 - cnt) * i) / 2;
> }
> for(int i = 1; i <= k / (1 + p); i++) ans += k % i;
> printf("%I64d\n", ans);
> }
> return 0;
> }
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