Question

用C语言编程：sum=1+2*3+4*5*6+7*8*9*10+…… 共n项，n由键盘输入

Answer 1

这个数很大，就算超大整数，也顶不过n=20，一般在15左右，还是能存放！

#include <stdio.h>

void main()
{
int i, j, n, k;
__int64 sum = 1, sum1=0;
scanf("%d", &n);
j = 1;
for(i = 1; i <= n; i++ )
{
sum = 1;
for (k=0;k<i; k++,j++)
{
sum *=j;
}
sum1 +=sum;
}

printf("sum1=%I64d\n", sum1);
}

另外写了两个大数相乘和相加的函数，这个n可以大到200.

#include <stdio.h>
#include <string.h>
#include <stdlib.h>
#define MAX 1000

//字符串逆置，思想：在字符串末尾分别设一指针
//两指针指向的值交换
char *ReverseStr(char *str)
{
char *p, *q, ch;
//去字符串长度
int len = strlen(str) - 1;
//当两指针没有相遇，指针的值就交换
for (p = str, q = str + len; p < q; p++,q--)
{
ch = *q;
*q = *p;
*p = ch;
}
//返回结果
return str;
}
//相乘
char *BigIntMulti(char *s1, char *s2)
{
int len, m = 0; //m字符串的位数
int len1 = strlen(s1);
int len2 = strlen(s2);
int flag = 0;
//取最大字符串长度
len = len1 > len2 ? len1: len2;
//给和字符串分配空间,长度暂时设成len1*len2
char *sum = (char *) malloc((len1 * len2 + 1) * sizeof(char));

//分配的空间初始化为0
memset(sum, '0', (len1 * len2 + 1) * sizeof(char));

//从字符串末尾进行处理
for(int i = len1 - 1; i >= 0; i--)
{
//m是积字符串的下标
m = len1 - 1 - i;
//对s2字符串也是从末尾进行处理,字符串'9'-'0'就成为整数9，所以按照乘法原则进行运算
for(int j = len2 - 1, temp = 0; j >=0; j--)
{
//temp保存 s1[i]*s2[j]的积和进位之和,sum[m]可能不为'0'，因为有进位，在之后的while循环之中设置的
temp = (sum[m] - '0') + (s1[i] - '0') * (s2[j] - '0') + flag;

//对temp的结果取余存入积字符串中
sum[m ++] = temp % 10 + '0';

//是否有进位
flag = temp / 10;
}
//最高一位可能有进位
while(flag)
{
//这里把进位加入到积字符串中
sum[m ++] += flag % 10;
//取余
flag /= 10;
}
}
//设置积字符串结束标记
sum[m] = '\0';
//返回结果，以上sum的结果是倒序，所以需要把逆置过来
return ReverseStr(sum);
}

//相加2,用指针处理
char *BigIntPlus(char *s1, char *s2)
{
int len, temp;
int len1 = strlen(s1);
int len2 = strlen(s2);
int flag = 0;
char *sum, *r;
len = len1 > len2 ? len1: len2;
sum = (char *) malloc((len + 2) * sizeof(char));
memset(sum, 0x00, (len + 2) * sizeof(char));
r = sum;
for(char *p = s1 + len1 - 1, *q = s2 + len2 - 1; (p >= s1 || q >= s2); p--, q--, r++)
{
if(p < s1)
temp = (*q - '0') + flag;
else if( q < s2)
temp = (*p - '0') + flag;
else
temp = (*p - '0' + *q - '0') + flag;

//把相加的结果添加到和字符串中
*r = temp % 10 + '0';

//保存相加结果的进位
flag = temp / 10;
}
//处理最后一位进位
if(flag) *r++ = flag % 10 + '0';
*r = '\0';
return ReverseStr(sum);
}

void main()
{
int i, j, n, k;
char sum[MAX] = {"1"}, sum1[MAX] ={"0"};
char tmp[MAX],tmp1[MAX],tmp2[MAX];

scanf("%d", &n);
j = 1;
for(i = 1; i <= n; i++ )
{
strcpy(sum, "1");
for (k = 0; k < i; k++, j++)
{
itoa(j, tmp, 10);
strcpy(tmp1, BigIntMulti(sum, tmp));
strcpy(sum, tmp1);
}
strcpy(tmp2, BigIntPlus(sum1, sum));
strcpy(sum1, tmp2);
}
printf("sum1=%s\n", sum1);
}

Answer 2

#include<stdio.h>
main()
{
int i,j,t,m,n,sum;
sum=0;
scanf("n=%d",&n);
for(i=0;i<=n,i++)
{ t=1;
for(j=1;j<=i;j++)
{ t+=j;
m*=t;}
sum+=m;
}
printf("sum=%d/n"sum);
}

Answer 3

#include <stdio.h>
void main()
{
int n,i;
long sum=0;
printf("please enter n:\t");
scanf("%d", &n);
for(i=1;i<=n;i++);
{
sum+=i*(i-1);
}
printf("\nsum=%ld\n", sum);
}

Answer 4

#include <stdio.h> //超长整数
__int64 main()
{
__int64 n,i,sum=0;
scanf("%I64d", &n);
for(i=1;i<=n;i++);
{
sum+=i*(i-1);
}
printf("%I64d\n", sum);
}

Answer 5

#include <stdio.h>
void main()
{
int n,i,a;
long sum=0;
a=0;
printf("please enter n:\t");
scanf("%d", &n);
for(i=1;i<=n);
{
sum+=i*(a+1);

i=i+1;
a=i;
}
printf("\nsum=%ld\n", sum);
}
呵呵，对不啊。

Answer 6

#include <stdio.h>
void main()
{
int n,i;
long sum=1; //sum=1
printf("please enter n:\t");
scanf("%d", &n);
for(i=1;i<=n;i++);
{
sum+=(i+1)*(i+2);
}
printf("\nsum=%ld\n", sum);
}