1) 读文件的内容(例如):
12
34
56
输出到:
56
34
12
(逆序)
2)输出和为一个给定整数的所有组合
例如n=5
5=1+4;5=2+3(相加的数不能重复)
则输出
1,4;2,3。
第一题,注意可增长数组的应用.
#include
#include
int main(void)
{ int MAX = 10;
int *a = (int *)malloc(MAX * sizeof(int));
int *b;
FILE *fp1;
FILE *fp2;
fp1 = fopen(“”,”r”);
if(fp1 == NULL)
{printf(“error1″);
exit(-1);
}
fp2 = fopen(“”,”w”);
if(fp2 == NULL)
{printf(“error2″);
exit(-1);
}
int i = 0;
int j = 0;
while(fscanf(fp1,”%d”,&a[i]) != EOF)
{i++;
j++;
if(i >= MAX)
{
MAX = 2 * MAX;
b = (int*)realloc(a,MAX * sizeof(int));
if(b == NULL)
{printf(“error3″);
exit(-1);
}a = b;
}}
for(;–j >= 0;)
fprintf(fp2,”%dn”,a[j]);
fclose(fp1);
fclose(fp2);
return 0;
}
第二题.
#include
int main(void)
{unsigned long int i,j,k;
printf(“please input the numbern”);
scanf(“%d”,&i);
if( i % 2 == 0)
j = i / 2;
else
j = i / 2 + 1;
printf(“The result is n”);
for(k = 0; k < j; k++)
printf(“%d = %d + %dn”,i,k,i – k);
return 0;
}
#include
void main()
{unsigned long int a,i=1;
scanf(“%d”,&a);
if(a%2==0)
{ for(i=1;i printf(“%d”,a,a-i);
}
else
for(i=1;i<=a/2;i++)
printf(” %d, %d”,i,a-i);
}
兄弟,这样的题目若是做不出来实在是有些不应该, 给你一个递规反向输出字符串的例子,可谓是反序的经典例程.
void inverse(char *p)
{ if( *p = = ‘