Pat 1069(同basic level里的1019)
2015-02-17 09:36
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#include<stdio.h> int main() { int a[4], n, b[4] = { 0, 0, 0, 0 }, i, j, temp; scanf("%d", &n); a[0] = n / 1000; a[1] = (n - a[0] * 1000) / 100; a[2] = (n - a[0] * 1000 - a[1] * 100) / 10; a[3] = n % 10; if (a[0] == a[1] && a[0] == a[2] && a[0] == a[3]) printf("%d - %d = 0000\n", n, n); else if (n == 6174) printf("7641 - 1467 = 6174\n"); else { while (n != 6174) { for (i = 0; i<4; i++) b[i] = a[i]; for (i = 0; i<4; i++) for (j = i + 1; j<4; j++) if (a[i]<a[j]){ temp = a[i]; a[i] = a[j]; a[j] = temp; } for (i = 0; i<4; i++) for (j = i + 1; j<4; j++) if (b[i]>b[j]) { temp = b[i]; b[i] = b[j]; b[j] = temp; } n = a[0] * 1000 + a[1] * 100 + a[2] * 10 + a[3] - (b[0] * 1000 + b[1] * 100 + b[2] * 10 + b[3]); printf("%d%d%d%d - %d%d%d%d = ", a[0], a[1], a[2], a[3], b[0], b[1], b[2], b[3]); a[0] = n / 1000; a[1] = (n - a[0] * 1000) / 100; a[2] = (n - a[0] * 1000 - a[1] * 100) / 10; a[3] = n % 10; printf("%d%d%d%d\n", a[0], a[1], a[2], a[3]); } } return 0; }
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