請(qǐng)編寫(xiě)一個(gè)函數(shù)fun(int*a,int n,int *odd,int *even),函數(shù)的功能是分別求出數(shù)組a中所有的奇數(shù)之和偶數(shù)之和
請(qǐng)編寫(xiě)一個(gè)函數(shù)fun(int*a,int n,int *odd,int *even),函數(shù)的功能是分別求出數(shù)組a中所有的奇數(shù)之和,通過(guò)even返回偶數(shù)之和.
例如：a數(shù)組元素中的值依次為：1,9,2,3,11,6；則通過(guò)odd返回奇數(shù)之和24；則通過(guò)even返回偶數(shù)之和8.
#include
#define N20
fun (int *a, int n, int *odd, int *even)
{
就是這里少點(diǎn)東西,大家?guī)兔?xiě)下
}
main( )
{ inta[N]={1,9,2,3,11,6},i, n=6, odd, even;
printf ( "The original data is : \n" );
for ( i = 0; i < n; i ++ ) printf ( "%5d", *(a+i) );
printf("\n\n");
fun ( a, n, &odd, &even );
printf ( "The sum of odd numbers:%d\n", odd );
printf ( "The sum of even numbers:%d\n", even );
NONO() ;
}
NONO()
{/* 本函數(shù)用于數(shù)據(jù)讀入和結(jié)果寫(xiě)入文件, 考生無(wú)需修改 */
int i, j, odd, even, n, a[N] ;
FILE *rf, *wf ;
rf = fopen("in.dat", "r") ;
if(rf == NULL) {
printf("在考生文件夾下數(shù)據(jù)文件in.dat不存在!") ;
return ;
}
wf = fopen("out.dat", "w") ;
for(i = 0 ; i < 10 ; i++) {
fscanf(rf, "%d", &n) ;
for(j = 0 ; j < n ; j++) fscanf(rf, "%d", a+j) ;
fun(a, n, &odd, &even) ;
fprintf(wf, "odd=%d,even=%d\n", odd, even) ;
}
fclose(rf) ;
fclose(wf) ;
}