C语言输入一个数组，再输入一个数N，删去这个数组中第N个数并打印数组
\u5728c\u8bed\u8a00\u4e2d\uff0c\u8f93\u5165\u4e00\u4e2a\u6570\u5b57n\uff0c\u5728\u4e0b\u4e00\u884c\u8f93\u5165n\u4e2a\u6570\u636e\uff0c\u7528\u6570\u7ec4\u3002#include int main(){ int n,a[100],b[100],c[100],i,j,k; scanf("%d",&n); for(i=0;i<n;i++) { scanf("%d",&a[i]); //\u591a\u4e86\u7a7a\u683c\uff0c\u4e0b\u540c } for(j=0;j<n;j++) { scanf("%d",&b[j]); } for(k=0;k<n;k++) { c[k]=a[k]+b[k]; printf("%d ",c[k]); } return 0;}
#include "stdio.h"
#define N 30
void main()
{
int a[N], i, j, k, n; n=N;
for(i=0; i<n; i++ )
scanf ( "%d", &a[i] ); //\u8fd9\u4e2a\u5faa\u73af\u662f\u4e3a\u4e86\u8f93\u5165\u6570\u7ec4\u5143\u7d20
for (i=0; i<n-1; i++ )
for (j = i+1; j< n; j++ ) //j=i+1
if (a[i]== a[j] )
{
for(k = j; k<n; k++ )
a[k] = a[k+1] ; /*\u540e\u9762\u7684\u6570\u5411\u524d\u79fb\u52a8 a[k]=a[k+1];*/
n--; /* \u5220\u9664\u540e\uff0c\u5c06\u6570\u7ec4\u5143\u7d20\u7684\u603b\u6570\u51cf1 n--;*/
j--;
}
for ( i=0; i<n; i++ )
printf("%8d", a[i] );
}
#include<stdio.h>
int main()
{
int i , a[100] , n , N ;
scanf("%d" , &n );
for( i = 0 ; i < n ; i++ )
scanf("%d" , &a[i]);
scanf("%d" , &N );
for( i = N - 1 ; i < n ; i++ )
{
a[i] = a[i+1] ;
}
a[n-1] = NULL ;
for( i = 0 ; i < n - 1 ; i++ )
printf("%d " , a[i] ) ;
return 0 ;
}
#include<stdio.h>
main()
{
doublo a[ ];
int N,m,j,i;
for(i=0;a[i]!='\0';i++)
{
scanf("%d",a);
m++;
}
scanf("%d",N);
free(a[N]);
for(j=N;N<m-1;j++,N++)
a[j]=a[N+1];
for(i=o;i<m-1;i++)
printf("%d",a[i]);
}
sadas
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