继续做一下分解质因数的水题！
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1 #include <cstdio> 2 #include <cstring> 3 #include <cstdlib> 4 #include <cmath> 5 #include <algorithm> 6 #include <iostream> 7 8 #define debug 0 9 10 using namespace std; 11 12 typedef __int64 ll; 13 const int maxn = 100005; 14 15 bool np[maxn]; 16 int pn, pr[maxn >> 2]; 17 18 void gp(){ 19 memset(np, 0, sizeof(np)); 20 np[0] = np[1] = true; 21 pn = 0; 22 for (int i = 2; i < maxn; i++){ 23 if (!np[i]) pr[pn++] = i; 24 for (int j = 0; j < pn && pr[j] * i < maxn; j++){ 25 np[pr[j] * i] = true; 26 if (i % pr[j] == 0) break; 27 } 28 } 29 #if debug 30 printf("pn %d\n", pn); 31 #endif 32 } 33 34 void fac(int a, int *f, int *n, int &cnt){ 35 int i = 0; 36 37 cnt = 0; 38 if (!a) return ; 39 while (pr[i] * pr[i] <= a && i < pn){ 40 if (a % pr[i] == 0){ 41 f[cnt] = pr[i]; 42 n[cnt] = 0; 43 while (a % pr[i] == 0) a /= pr[i], n[cnt]++; 44 cnt++; 45 } 46 i++; 47 } 48 if (a != 1) f[cnt] = a, n[cnt++] = 1; 49 } 50 51 int cnt_fac(int n, int f){ 52 int ep = f; 53 int ret = 0; 54 55 while (ep <= n){ 56 ret += n / ep; 57 ep *= f; 58 } 59 60 return ret; 61 } 62 63 bool deal(int a, int b){ 64 int f[30], cf[30]; 65 int n = 0; 66 67 memset(f, 0, sizeof(f)); 68 fac(a, f, cf, n); 69 #if debug 70 printf("n %d\n\n", n); 71 for (int i = 0; i < n; i++) { 72 printf("%d %d\n", f[i], cf[i]); 73 } 74 f[n] = 0; 75 #endif 76 77 for (int i = 0; i < n; i++) { 78 #if debug 79 printf("fac %d num %d\n", f[i], cnt_fac(b, f[i])); 80 #endif 81 if (cf[i] > cnt_fac(b, f[i])) { 82 return false; 83 } 84 } 85 return true; 86 } 87 88 int main(){ 89 int a, b; 90 91 gp(); 92 while (~scanf("%d%d", &a, &b)){ 93 if (deal(b, a) && b) { 94 printf("%d divides %d!\n", b, a); 95 } 96 else { 97 printf("%d does not divide %d!\n", b, a); 98 } 99 } 100 101 return 0; 102 }
——written by Lyon