【Codeforces Round #268 (Div. 1)】A. 24 Game【归纳法】
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首先我们知道小于4的输入都是肯定不能构成24的,我假设输入为4到k都是可行的,则证明k+1是可行的。
若输入是k+1,则我们先让k+1 - k = 1.然后就只剩下1~k-1和一个1这么多数字,我们已经知道k-1是可行的,然后k-1求得的24再乘上最后一个1即可得到答案。所以用归纳法可以做出这题。
#include <cstdio>#include <cstring>void cal(int n) { if (n == 4) { printf("1 * 2 = 2\n2 * 3 = 6\n6 * 4 = 24\n"); }else if (n == 5) { printf("5 - 2 = 3\n3 + 3 = 6\n6 * 1 = 6\n6 * 4 = 24\n"); }else { printf("%d - %d = 1\n", n, n-1); cal(n-2); printf("1 * 24 = 24\n"); }}int main() { int n; while (~scanf("%d", &n)) { if (n < 4) puts("NO"); else { puts("YES"); cal(n); } }}
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