如何计算从1!到n!的所有阶乘值？

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c

1、求阶乘乘+n：

c#include

int main() { int i, n, ret; printf(输入n的值：); scanf(%d, &n); ret=1; for (i=1; i <=n; i++) { ret *=i; } printf(n!=%d\n, ret); return 0;}

2、求阶乘之和：c#include

int main() { int i, j, n, ret, sum=0; printf(输入n的值：); scanf(%d, &n); for (i=1; i <=n; i++) { ret=1; for (j=1; j <=i; j++) { ret *=j; } sum +=ret; } printf(阶乘之和=%d\n, sum); return 0;}

1、求阶乘 n！

int main() { int i, n, ret; printf("输入n的值："); scanf("%d", &n); ret = 1; for (i = 1; i <= n; i++) { ret = ret * i; } printf("n!= %d",ret); return 0; }

2、求阶乘之和 n!+(n-1)!+...+1!

int main() { int i, j, n, ret, sum; printf("输入的n是："); scanf("%d", &n); sum = 0; for (j = 1; j <= n; j++) { ret = 1; for (i = 1; i <= j; i++) { ret = ret * i; } sum = sum + ret; } printf("n!= %d\n", ret); printf("n!+(n-1)!+...+1!= %d\n", sum); return 0; }

性能优化后代码，

int main() { int i, n, ret, sum; printf("输入的n是："); scanf("%d", &n); ret = 1; sum = 0; for (i = 1; i <= n; i++) { ret = ret * i; sum = sum + ret; } printf("n!= %d\n", ret); printf("n!+(n-1)!+...+1!= %d\n", sum); return 0; }

输出结果：