ACM/ICPC World Finals 2013 D Factors
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题目大意
定义
解答
首先将k分解质因数:
很容易就可以得到:
对于一个新加的素数:
然后我们就可以发现我们只要取前若干个质数就可以了,然后我们暴力维护每个n所对应的k(
#include <cstdio>#include <cstdlib>#include <cmath>#include <cstring>#include <algorithm>#include <map>using namespace std;#define LLU unsigned long long intconst LLU oo = 1ULL<<63;LLU yh[100][100];LLU ss[50] = {2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67};map<LLU, LLU> anses;//求组合数void preWork(){ yh[0][0] = 1; for (int i = 1; i <= 62; i++) { yh[i][0] = 1; for (int j = 1; j <= 62; j++) { if (yh[i-1][j] <= oo && yh[i-1][j-1] <= oo) yh[i][j] = yh[i-1][j] + yh[i-1][j-1]; else yh[i][j] = oo; } }}//暴力搜索维护答案void work(LLU k, LLU n, LLU count, LLU i){ if (count && (!anses.count(k) || anses[k] > n)) anses[k] = n; LLU tot = 0; LLU tmp = n; while(ss[i] <= oo/tmp) { tot++; tmp *= ss[i]; work(k*yh[count+tot][tot], tmp, count+tot, i+1); }}int main(){ //freopen("factors.in", "r", stdin); //freopen("factors.out", "w", stdout); preWork(); work(1, 1, 0, 0); LLU x; while (scanf("%I64u", &x) != EOF) { printf("%I64u %I64u\n", x, anses[x]); } return 0;}
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