題意：給定m和n，求出最大的k，使得m^k能整除n的階乘。

思路：各自分解質(zhì)因數(shù)，然后對比下就能找出k值。

# include <iostream> # include <cstdio> # include <cstring> using namespace std; const int MAXN = 10000; int a[MAXN+1]={0}, b[MAXN+1], c[MAXN][2], d[200][2], k=0;void init() {for(int i=2; i<=MAXN; ++i)if(!a[i]){b[k++] = i;for(int j=i; j<=MAXN; j+=i)a[j] = 1;} }void fun1(int n, int &cnt)//分解n {int tmp = n;for(int i=0; i<k; ++i){if(n >= b[i]){c[cnt][0] = b[i];while(tmp){c[cnt][1] += tmp/b[i];tmp /= b[i];}tmp = n;++cnt;}elsebreak;}if(!a[n]){c[cnt][0] = n;c[cnt][1] = 1;++cnt;} }void fun2(int n, int &cnt)//分解m {for(int i=0; i<k && b[i]*b[i]<=n; ++i){if(n % b[i] == 0){d[cnt][0] = b[i];for(;n%b[i]==0;d[cnt][1]++,n/=b[i]);++cnt;}}if(n != 1){d[cnt][0] = n;d[cnt][1] = 1;++cnt;} }int main() {int t, m, n, cnt1, cnt2, ans,cas=1;init();scanf("%d",&t);while(t--){cnt1 = cnt2 = 0;ans = 0x3f3f3f3f;bool flag = true;memset(c, 0, sizeof(c));memset(d, 0, sizeof(d));scanf("%d%d",&m,&n);printf("Case %d:\n",cas++);fun1(n, cnt1);fun2(m, cnt2);for(int i=0; i<cnt2; ++i){bool ok = false;for(int j=0; j<cnt1; ++j){if(c[j][0] == d[i][0]){if(c[j][1] >= d[i][1]){ok = true;ans = min(ans, c[j][1]/d[i][1]);}elsebreak;}}if(!ok){flag = false;break;}}if(!flag)puts("Impossible to divide");elseprintf("%d\n",ans);}return 0; }

轉(zhuǎn)載于:https://www.cnblogs.com/junior19/p/6729975.html

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