#include <stdio.h>
void main(){
    int n,a[5];
	scanf("%d",&n);
    while(n>=0&&n<=20){
        for(int j=1;j<=n;j++){
            for(int i=0;i<4;i++)
	            scanf("%d",&a[i]);
			if((a[0]+a[2]==2*a[1])&&(a[1]+a[3]==2*a[2]))
				a[4]=2*a[3]-a[2];
			else if((a[0]*a[2]==a[1]*a[1])&&(a[1]*a[3]==a[2]*a[2])&&(a[1]%a[0]==0))
				a[4]=a[3]*a[3]/a[2];
			else break;
            for(i=0;i<=4;i++)
	            printf("%d ",a[i]);
			    printf("\n");
		}
	}
}

• Re:output limited exceed (391) LIBAI 2004-12-23 18:08:27
• Re:output limited exceed (706) LIBAI 2004-12-26 10:40:47