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110ms水过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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