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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; > } Followed by: Post your reply here: |
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