下面程序可求出矩阵两条对角线上的元素之和。请分析程序填空。main(){int a[3][3]= {1,3,6,7,9,11,14,15,17},sum1=0,sum2=0,i,j;for(i=0; i<3; i++)for(j=0; j<3; j++)if(i==j) sum1+=a[i][j];for(i=0; i<3; i++)for(【1】; 【2】; j--)if(i+j==2) sum2+=a[i][j];printf("sum1=%d,sum2=%d\n",sum1,sum2);}
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下面程序的运行结果是() 。main(){int a[5][5],i,j,n=1;for(i=0; i<5; i++)for(j=0; j<5; j++)a[i][j]=n++;printf("The result is:\n");for(i=0; i<5; i++){for(j=0; j<=i; j++)printf("%4d",a[i][j]);printf("\n");}}
若有以下输入,则下面程序的运行结果是()。1 4 2 3 3 4 1 2 3 3 2 2 2 3 3 1 1 1 4 1 1 1 -1#define M 50main(){int a[M],c[4],i,n=0,x;printf("Enter 1 or 2 or 3 or 4,to end with -1\n");scanf("%d",&x);while(x!=-1){if(x>0&&x<=4){a[n-1]=x;n++;}scanf("%d",&x);}for(i=0; i<4; i++) c[i]=0;for(i=0; i
下面程序是将十进制数转换成不超过十六进制的n进制。请分析程序填空。#includemain(){int i=0,base,n,j,num[20];printf("Enter data that will be converted:\n");scanf("%d",&n);printf("Enter base:\n");scanf("%d",&base);do{i++;num[i]=n【1】base;n=n【2】base;}while(n);printf("The data %d has been converted into the %d base data:\n",n,base);for(【3】)if(num[j]>9) putchar(num[j]+'A'-10);else putchar(num[j]+'0');}
下面程序的功能是输入5个整数,找出最大数和最小数所在的位置,并把二者对调,然后输出调整后的5个数。main(){int a[5],max,min,i,j,k;for (i=0; i<5; i++)scanf("%d:,&a[i];min=a[0];for (i=1;i<5;i++)if (a[i]max) {max=a[i]; 【2】;}【3】printf("\nThe position of min is:%3d\n",k);printf("\nThe position of max is:%3d\n",j);for (i=0;i<5;i++)printf("%5d",a[i]);}