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这个是BZOJ1005的简化版
因为不会爆long long所以不需要高精度了,但是如果直接计算,中间过程会爆long long
一个很好的方法就是分解质因数了,最后算完,把所有因子都乘起来
1 /************************************************************** 2 Problem: 1211 3 User: round_0 4 Language: C++ 5 Result: Accepted 6 Time:0 ms 7 Memory:1276 kb 8 ****************************************************************/ 9 10 #include <iostream> 11 #include <cstdio> 12 using namespace std; 13 const int maxn = 305; 14 typedef long long LL; 15 int a[maxn] = {0}; 16 int fac[maxn] = {0}; 17 int n; 18 void f(int x,int b){ 19 while(x>1){ 20 for(int j = 2;j<=n;++j)if(x%j==0){ 21 fac[j]+=b; 22 x/=j; 23 break; 24 } 25 } 26 } 27 int main() 28 { 29 // freopen("in.txt","r",stdin); 30 scanf("%d",&n); 31 int s = 0; 32 for(int i = 1;i<=n;++i)scanf("%d",&a[i]); 33 if(n==1){ 34 if(a[1]!=0)printf("0\n"); 35 else printf("1\n"); 36 return 0; 37 } 38 for(int i = 1;i<=n;++i){ 39 if(a[i]<1){printf("0\n");return 0;} 40 a[i]--;s+=a[i]; 41 } 42 if(s!=n-2){printf("0\n");return 0;} 43 for(int i = 2;i<=n-2;++i)f(i,1); 44 for(int i = 1;i<=n;++i) 45 for(int k = 2;k<=a[i];++k) 46 f(k,-1); 47 LL ans = 1; 48 for(int i = 1;i<=n;++i) 49 for(int j = 1;j<=fac[i];++j) 50 ans*=i; 51 printf("%lld\n",ans); 52 return 0; 53 }
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原文地址:http://www.cnblogs.com/GJKACAC/p/4732411.html