2017 Multi-University Training Contest
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题目链接
输入二维字符数组,让判断上面显示的时间。
观察数字特征:
如果上面没有横,这样的数字只有1,4,很容易辨别1,4的特征,剩余的数字为0,2,3,5,6,7,8,9
然后看一判断中间没有横的数字,这样的数字只有0,7,然后可以再辨别这两个数的特征,剩余数字2,3,5,6,8,9
然后发现2这个数字,没有右下角那一竖,所以可以把2辨别出来,剩余3,5,6,8,9
然后剩余数字数,只有3左上角那一竖不存在,所以可以把3辨别出来,剩余5,6,8,9
而5,6,都没有右上角那一竖,可以再把5,6辨别出来。
剩余8,9很好辨别。
#include <iostream>#include <stdio.h>using namespace std;char g[10][30];int judge(int num) ///判断那个数字{ int L; if(num == 1) {L = 0;} if(num == 2) {L = 5;} if(num == 3) {L = 12;} if(num == 4) {L = 17;} if(g[1][L+1] != 'X') ///这样的数字有1,和4 { if(g[2][L] == 'X') return 4; else return 1; } else if(g[4][L+1] != 'X') ///这样的数字有0和7 { if(g[2][L] == 'X') return 0; else return 7; } else if(g[5][L+3] != 'X') { return 2; } else if(g[2][L] != 'X') { return 3; } else if(g[2][L+3] != 'X') ///这样的数有5,6 { if(g[5][L] == 'X') return 6; else return 5; } else { if(g[5][L] == 'X') return 8; else return 9; }}int main(){ int t; scanf("%d",&t); while(t--) { for(int i = 1; i <= 7; i++) { scanf("%s",g[i]); } int time1 = judge(1); int time2 = judge(2); int time3 = judge(3); int time4 = judge(4); printf("%d%d:%d%d\n",time1,time2,time3,time4); } return 0;}
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- 2017 Multi-University Training Contest
- 2017 Multi-University Training Contest
- 2017 Multi-University Training Contest
- 2017 Multi-University Training Contest
- 2017 Multi-University Training Contest
- 2017 Multi-University Training Contest
- 2017 Multi-University Training Contest
- 2017 Multi-University Training Contest
- 2017 Multi-University Training Contest
- 2017 Multi-University Training Contest
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- #2017 Multi-University Training Contest
- 2017 Multi-University Training Contest
- #2017 Multi-University Training Contest
- 2017 Multi-University Training Contest
- 2017 Multi-University Training Contest
- 2017 Multi-University Training Contest
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