N!

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 262144/262144 K (Java/Others) Total Submission(s): 75303    Accepted Submission(s): 21981 Problem Description Given an integer N(0 ≤ N ≤ 10000), your task is to calculate N!   Input One N in one line, process to the end of file.   Output For each N, output N! in one line.   Sample Input 1 2 3   Sample Output 1 2 6   Author 求n的阶乘嘛！ 基础题，依然WA，也是够了！ #include<iostream> #include<stdio.h> #include<string.h> #include<math.h> #include<algorithm> #include<stdlib.h> #include<queue> typedef long long ll; using namespace std; #define INF 0x3f3f3f3f #define N 1100000 int a[N]; int main() { int i,j,n,count,t,k; while(scanf("%d",&n)!=EOF) { a[0]=1; count=1; for(i=1;i<=n;i++) { k=0; for(j=0;j<count;j++) { t=a[j]*i+k; a[j]=t%10; k=t/10; } while(k) { a[count++]=k%10; k=k/10; } } for(i=count-1;i>=0;i--) { printf("%d",a[i]); } printf("\n"); } return 0; } 10000的阶乘也就35660位，要是担心TLE，或者超限的话，也可以按10000进制来写： #include<iostream> #include<stdio.h> #include<string.h> #include<math.h> #include<algorithm> #include<stdlib.h> #include<queue> typedef long long ll; using namespace std; #define INF 0x3f3f3f3f #define N 1100000 int a[N]; int main() { int i,j,n,count,t,k; while(scanf("%d",&n)!=EOF) { a[0]=1; count=1; for(i=1;i<=n;i++) { k=0; for(j=0;j<count;j++) { t=a[j]*i+k; a[j]=t%10000; k=t/10000; } while(k) { a[count++]=k%10000; ///这里以10000为进位单位 k=k/10000; } } printf("%d",a[count-1]); ///注意一下，第一个数不是“%04d”； for(i=count-2;i>=0;i--) { printf("%04d",a[i]); ///其他的按四位的输出，不足补零 } printf("\n"); } return 0; }