f(i*i%10=i) else if (i *k i% 100== i) 【题6.107】# include< stdio.h> mal printf("input x and y: " (“%6d%d”,&x,&y) for (i=l:i<=y:i++ last last *x% 1000 printf(\ The last 3 digits of %d **%d is: %d\n", x, y, last%1000) 【题6.108】# include< stdio.h> float g, sum, ave for(i=1;i<=6;i++) sum =0 f scanf(%f”,&g) ave sum/5 printf(No %d ave=%65 2f\n" 键盘输入:9570736764<CR 6065778690<CR 10093899285<CR> 5263667071<CR> 运行结果:No.1ave=73.80 No.2ave=75.60 No.3ave=79.40{ int i; for (i=1; i<100; i++) if (i * i % 10 == i) printf(“%3d”, i); else if (i * i % 100 == i) printf(“%3d”, i); } 【题 6.107】 #include <stdio.h> main() { int i, x, y, last = 1; printf(“Input x and y:”); scanf(“%d%d”, &x, &y); for (i=1; i<=y; i++) last = last * x % 1000; printf(“\nThe last 3 digits of %d * * %d is: %d\n”, x, y, last%1000); } 【题 6.108】 #include <stdio.h> main() { int i, j; float g, sum, ave; for (i=1; i<=6; i++) { sum = 0; for (j=1; j<=5; j++) { scanf(“%f”, &g); sum += g; } ave = sum / 5; printf(“No.%d ave=%5.2f\n”, i, ave); } } 键盘输入： 95 70 73 67 64<CR> 60 65 77 86 90<CR> 71 93 63 80 90<CR> 87 83 91 95 89<CR> 100 93 89 92 85<CR> 52 63 66 70 71<CR> 运行结果： No.1 ave=73.80 No.2 ave=75.60 No.3 ave=79.40