UVA 1363 - Joseph's Problem(数论)
2014-07-05 00:01
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UVA 1363 - Joseph's Problem
题目链接题意:给定n, k,求出∑ni=1(k mod i)
思路:由于n和k都很大,直接暴力是行不通的,然后在纸上画了一些情况,就发现其实对于k/i相同的那些项是形成等差数列的,于是就可以把整个序列进行拆分成[k,k/2],[k/2,
k/3], [k/3,k/4]...k[k/a, k/b]这样的等差数列,利用大步小步算法思想,这里a枚举到sqrt(k)就可以了,这样就还剩下[1,k/a]的序列需要去枚举,总时间复杂度为O(sqrt(k)),然后注意对于n大于k的情况,n超过k的部分全是等于k,为(n - k) * k,这样把所有部分加起来就是答案
这有一篇比较详细的题解
代码:
#include <stdio.h>
#include <string.h>
#include <math.h>
long long n, k;
long long solve() {
long long ans = 0;
if (n > k) ans += (n - k) * k;
long long a = (long long )sqrt(k), b = k / a;
for (long long i = a; i > 1; i--) {
long long a0 = k / i, an = k / (i - 1);
if (a0 > n) break;
if (an > n) an = n;
ans += (k % an + k % (a0 + 1)) * (an - a0) / 2;
}
for (int i = 1; i <= n && i <= b; i++) ans += k % i;
return ans;
}
int main() {
while (~scanf("%lld%lld", &n, &k)) {
printf("%lld\n", solve());
}
return 0;
}
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