C程序训练：阶乘与溢出-编程知识

C程序训练：阶乘与溢出

已知n是整数，计算1+2!+3!+...+n!，并给出最大能够计算的n值是多少？

1. 假设n是int类型，系统用32位表示int类型。代码如下：

#include <stdio.h>
int main()
{int n,sum1,sum=1,fact=1;int step;for(n=2; n<100; n++) {step = fact;fact *= n;if( step>fact) break;sum1 = sum;sum += fact;    if (sum1>sum) {sum = sum1;break;}}printf("sum=%d\n", sum);  printf("n=%d\n", n-1);  return 0;
}

程序运行结果如下图所示。

2. 假设n是long long int类型，即64位整数。代码如下：


#include <stdio.h>
int main()
{long long int sum1,sum=1,fact=1;long long int step;int n;for(n=2; n<100; n++) {step = fact;fact *= n;if( step>fact) break;sum1 = sum;sum += fact;    if (sum1>sum) {sum = sum1;break;}}printf("sum=%lld\n", sum);  printf("n=%d\n", n-1);  return 0;
}

程序运行结果如下图所示。

3. 假设n是unsigned int类型，即32位无符号整数。代码如下：


#include <stdio.h>
int main()
{unsigned int n,sum1,sum=1,fact=1;unsigned int step;for(n=2; n<100; n++) {step = fact;fact *= n;if( step>fact) break;sum1 = sum;sum += fact;    if (sum1>sum) {sum = sum1;break;}}printf("sum=%u\n", sum);  printf("n=%d\n", n-1);  printf("%d!=%u\n", n-1, step);  return 0;
}

程序运行结果如下图所示。

4. 假设n是unsigned long long int类型，即64位无符号整数。代码如下：


#include <stdio.h>
int main()
{unsigned long long int sum1,sum=1,fact=1;unsigned long long int step;int n;for(n=2; n<100; n++) {step = fact;fact *= n;if( step>fact) break;sum1 = sum;sum += fact;    if (sum1>sum) {sum = sum1;break;}}printf("sum=%llu\n", sum);  printf("n=%d\n", n-1);  printf("%d!=%llu\n", n-1, step);  return 0;return 0;
}

程序运行结果如下图所示。