关于for循环嵌套的问题

[syn724]

我是一名c语言初学者，在一本书上看见关于for循环嵌套的一篇代码，但是其中的语句有些问题不能编译，请大家帮我改进下，谢谢！
#include <stdio.h>

int main(void)
{
long sum = 1L;
int j = 1;
int count = 0;

/* Prompt for, and read the input count */
printf("\nEnter the number of integers you want to sum: ");
scanf(" %d", &count);

for(int i = 1 ; i <= count ; i++)
{
sum = 1L;
j=1;
printf("\n1");

/* Calculate sum of integers from 1 to i */
while(j < i)
{
sum += ++j;
printf("+%d", j);
}
printf(" = %ld\n", sum);
}
return 0;
}
书上面的程序输出如下：
Enter the number of integers you want to sum:5

1=1
1+2=3
1+2+3=6
1+2+3+4=10
1+2+3+4+5=15

[syn724]

呵呵 ，谢谢你。问题已经解决。不过我奇怪的是，书上的for循环的控制表达式中，都是在表达式里面定义的类型。但是这样写法在VC里面会出现报错。看来还是我RP的问题。

[wahl1986]

#include<iostream>
using namespace std;
int main()
{int n;
int sum=0;
cout<<"Enter the number of integers you wants to sum:";
cin>>n;
cout<<endl;
for(int i=1;i<=n;i++)
{sum=sum+i;
for(int j=1;j<i;j++)
{
cout<<j;
cout<<"+";
}

cout<<i<<"="<<sum;
cout<<endl;
}
return 0;
}比书上的简单

[artman]

ALL OK! with VC6

[烂泥]

#include <stdio.h> 

int main(void) 
{ 
  long sum = 1L;                      
  int j = 1;                         
  int count = 0;                      
  int i;
  /* Prompt for, and read the input count */ 
  printf("\nEnter the number of integers you want to sum: "); 
  scanf(" %d", &count); 

  for( i = 1 ; i <= count ; i++) 
  { 
    sum = 1L;                          
    j=1;                              
    printf("\n1"); 

    /* Calculate sum of integers from 1 to i */ 
    while(j < i) 
    { 
      sum += ++j; 
      printf("+%d", j);                
    } 
    printf(" = %ld\n", sum);          
  } 
  return 0; 

这样就行 了！先把i定义为整型！

[烂泥]

#include <stdio.h> 

int main(void) 
{ 
  long sum = 1L;                      
  int j = 1;                          
  int count = 0; 
  int i;                   

  /* Prompt for, and read the input count */ 
  printf("\nEnter the number of integers you want to sum: "); 
  scanf(" %d", &count); 

  for( i = 1 ; i <= count ; i++) 
  { 
    sum = 1L;                          
    j=1;                              
    printf("\n1"); 

    /* Calculate sum of integers from 1 to i */ 
    while(j < i) 
    { 
      sum += ++j; 
      printf("+%d", j);                
    } 
    printf(" = %ld\n", sum);          
  } 
  return 0; 
} 

这样！