【Codeforces Round #268 (Div. 1)】A. 24 Game【归纳法】
2014-09-27 10:41
417 查看
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。
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); } } }
相关文章推荐
- Codeforces Round#268(Div 2)C 24 Game
- Codeforces Round#268(Div 2)C 24 Game
- Codeforces Round #268 (Div. 2) C 24 Game
- Codeforces Round #268 (Div. 1) A. 24 Game 构造
- Codeforces Round #268 (Div. 1)A. 24 Game
- Codeforces Round #268 (Div. 1)A. 24 Game
- Codeforces Round #268 (Div. 2) C 24 Game [构造]
- Codeforces Round #268 (Div. 2)+24 Game
- D. String Game - Codeforces Round #402 (Div. 2)
- 【Codeforces Round #402 (Div. 1)】Codeforces 778A String Game
- Codeforces Round #426 (Div. 2) C. The Meaningless Game
- 【codeforces】 468A 24 Game
- Codeforces Round #402 (Div. 1) A. String Game
- CodeForces 468A A. 24 Game
- Codeforces Round #267 (Div. 2) B. Fedor and New Game
- Codeforces Round #280 (Div. 2) D. Vanya and Computer Game
- 【Codeforces Round #426 (Div. 2) C】The Meaningless Game
- Making a mobile game in 24 hours
- Codeforces Round #419 (Div. 2) C.Karen and Game 思维
- Code forces #258 div2 A.Game With Sticks