C - 基础 Homework
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//1. (**)随机产生20个[10 , 100]的正整数,输出这些数以及他们中的最大数 { int max = 10, num; printf("随机产生20个[10 , 100]的正整数:"); for (int i = 1; i <= 20; i++) { num = arc4random() % (1000 - 10 + 1) + 10; printf("%d ", num); max = num > max ? num : max; } printf("\n最大数为:%d\n", max); }//2. (**)编程将所有“水仙花数”打印出来,并打印其总个数。 “水仙花数”是一个 各个位立方之和等于该整数的三位数。 { int count, a, b, c; for (int i = 100; i < 1000; i++) { a = i / 100; b = i % 100/ 10; c = i % 10; if (i == a * a * a + b * b * b + c * c * c) { printf("%d ",i); } count++; } }//3. (**)已知abc+cba = 1333,其中a,b,c均为一位数,编程求出满足条件的a,b,c所有组合 { printf( "a,b,c所有组合为:"); for (int a = 0; a < 10; a++) { for (int b = 0; b < 10; b++) { for (int c = 0; c < 10; c++) { if(a * 100 + b * 10 + c + c * 100 + b * 10 + a == 1333){ printf("%d%d%d ",a ,b ,c); } } } } printf("\n"); }//4. (***)输入两个数,求最大公约数和最小公倍数。(用两种方法:辗转相除法和普通方法)//求最大公约数,辗转相除法: { int i, j, a, b, count, m; printf("请输入两个数:"); scanf("%d%d",&a,&b); i = a; j = b; while (i % j != 0) { m = i % j; i = j; j = m; count++; } printf("最大公约数为:%d,计算次数为:%d。\n", j, count); printf("最小公倍数为:%d\n",a * b / j); }//普通方法: { int a, b, i; printf("请输入两个数:"); scanf("%d%d",&a,&b); int num = a >= b ? b : a; for ( i = num ; i > 0; i--) { if (a % i == 0 && b % i == 0) { printf("最大公约数为:%d\n",i); break; } } printf("最小公倍数为:%d\n",a * b / i); }//5. (***)一个球从100m高度自由落下,每次落地后反跳回原来高度的一半,再落下,再反弹。求它在第10次落地时,共经过多少米?第10次反弹多高 { float high = 100,sum = 100;int down = 0; while (down < 10) { high = high / 2; sum += high * 2; down ++; } sum -= high*2; printf("第10次落地时,共经过%.2f米\n",sum); printf("第10次反弹高度为:%.2f米\n",high); }//6. (****)输入n,分别用*输出边长为n的实心菱形和空心菱形。// 实心菱形: { int n, j, m, num; printf("请输入一个数:"); scanf("%d",&n); for (int count = 1; count <= 2 * n - 1; count++) { num = (count > n ) ? (2 * n - count ): count ; j = n - num;//空格 m = 2 * n - 1 - j * 2;//星星 for (int i = 1 ; i <= j ; i++ ) { printf(" "); } for (int i = 1; i <= m; i++) { printf("*"); } printf("\n"); } }// 空心菱形: { int n, j, m, num ; printf("请输入一个数:"); scanf("%d",&n); for (int count = 1; count <= 2 * n - 1; count++) { num = (count > n ) ? (2 * n - count): count ; j = n - num;//空格 m = 2 * n - 1 - j * 2;//星星 for (int i = 1 ; i <= j ; i++ ) { printf(" "); } for (int i = 1; i <= m; i++) { if (i == 1 || i == m) { printf("*"); } else{ printf(" "); } } printf("\n"); } }
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