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Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 7943 Accepted Submission(s): 4261
#include <stdio.h> #include <iostream> using namespace std; int a[105][100]; void ktl(){ int i,j,yu,len; a[2][0]=1; a[2][1]=2; a[1][0]=1; a[1][1]=1; len =1; for(i=3;i<101;i++) { yu=0; for(j=1;j<=len;j++) { int t=(a[i-1][j])*(4*i-2)+yu; yu=t/10; a[i][j]=t%10; } while(yu) { a[i][++len]=yu%10; yu/=10; } for(j=len;j>=1;j--) { int t=a[i][j]+yu*10; a[i][j]=t/(i+1); yu = t%(i+1); } while(!a[i][len]) { len--; } a[i][0]=len; } } int main() { /* code */ int n; ktl(); while(~scanf("%d",&n)){ for(int i=a[n][0];i>0;i--){ printf("%d",a[n][i] ); } printf("\n"); } return 0; }
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原文地址:http://www.cnblogs.com/superxuezhazha/p/5564471.html