396A. On Number of Decompositions into Multipliers
题目链接:
http://codeforces.com/problemset/problem/396/A
题目意思:
有n(n<=500)个数ai(ai<=10^9),令m=a1*a2*..*an 求把m划分成n个数相乘的划分种数。
解题思路:
分解质因数+组合数学,首先对每个数分解质因数,最后求得m=a1^k1*a2^k2*a3^k3*...*ap*kp
依次把每个质因数放进这n个数框中,要凑成n个数,构造n-1个相同的隔板,然后将ai和n-1个隔板排列,排列后按隔板位置分开,没有的用1来代替,最后可以把该质因数放进n个数框中的种数求出来,连乘后即可得出答案。
代码:
//#include<CSpreadSheet.h> #include<iostream> #include<cmath> #include<cstdio> #include<sstream> #include<cstdlib> #include<string> #include<string.h> #include<cstring> #include<algorithm> #include<vector> #include<map> #include<set> #include<stack> #include<list> #include<queue> #include<ctime> #include<bitset> #define eps 1e-6 #define INF 0x3f3f3f3f #define PI acos(-1.0) #define ll __int64 #define LL long long #define lson l,m,(rt<<1) #define rson m+1,r,(rt<<1)|1 #define M 1000000007 //#pragma comment(linker, "/STACK:1024000000,1024000000") using namespace std; #define Maxn 550 ll c[22000][Maxn]; void init() { memset(c,0,sizeof(c)); c[1][0]=1; c[1][1]=1; for(int i=2;i<=20000;i++) //32*500 { c[i][0]=1; for(int j=1;j<=500&&j<=i;j++) c[i][j]=(c[i-1][j-1]+c[i-1][j])%M; //求出组合数 } } int main() { //freopen("in.txt","r",stdin); //freopen("out.txt","w",stdout); init(); int n; while(~scanf("%d",&n)) { map<int,int>myp; for(int i=1;i<=n;i++) { ll temp; scanf("%I64d",&temp); for(ll j=2;j*j<=temp;j++) //分解质因数 { if(temp%j==0) { while(temp%j==0) { myp[j]++; temp/=j; } } } if(temp>1) myp[temp]++; } ll ans=1; map<int,int>::iterator it=myp.begin(); for(;it!=myp.end();it++) //分别对每个质因数处理 { // printf("%d %d\n",it->second,c[it->second+n-1][n-1]); ans=(ans*c[it->second+n-1][n-1])%M; } printf("%I64d\n",ans); } return 0; }
原文:http://blog.csdn.net/cc_again/article/details/20037441