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[1219] Time
- 时间限制: 1000 ms 内存限制: 131072 K
- 问题描述
- Digital
clock use 4 digits to express time, each digit is described by 3*3 characters (including”|”,”_”and” “).now given the current time, please tell us how can it be expressed by the digital clock. - 输入
-
There are several test cases.
Each case contains 4 integers in a line, separated by space.
Proceed to the end of file.
- 输出
- For each test case, output the time expressed by the digital clock such as Sample Output.
- 样例输入
-
1 2 5 6 2 3 4 2
- 样例输出
-
_ _ _ | _||_ |_ ||_ _||_| _ _ _ _| _||_| _| |_ _| ||_
- 提示
-
The digits showed by the digital clock are as follows: _ _ _ _ _ _ _ _ | _| _||_||_ |_ ||_||_|| | ||_ _| | _||_| ||_| _||_|
- 来源
-
辽宁省赛2010
题目意思也很简单就是用字符串模拟时钟，注意每个数字占三格，最暴力的方法就是直接定义三个字符数组，按要求输出，还有一点就是0在最后一位，要特殊处理一下。
#include <algorithm> #include <iostream> #include <cstring> #include <cstdlib> #include <cstdio> #include <string> #include <vector> #include <cmath> #include <ctime> #include <queue> #include <stack> #include <set> #include <map> using namespace std; typedef long long LL; const int maxn= 100000 + 10; char s1[35]=" _ _ _ _ _ _ _ _ "; char s2[35]=" | _| _||_||_ |_ ||_||_|| |"; char s3[35]=" ||_ _| | _||_| ||_| _||_|"; int main() { int a,b,c,d; while(~scanf("%d%d%d%d",&a,&b,&c,&d)) { getchar(); if(a==0) a=10; if(b==0) b=10; if(c==0) c=10; if(d==0) d=10; printf("%c%c%c%c%c%c%c%c%c%c%c%c\n",s1[a*3-2],s1[a*3-1],s1[a*3],s1[b*3-2],s1[b*3-1],s1[b*3],s1[c*3-2],s1[c*3-1],s1[c*3],s1[d*3-2],s1[d*3-1],s1[d*3]); printf("%c%c%c%c%c%c%c%c%c%c%c%c\n",s2[a*3-2],s2[a*3-1],s2[a*3],s2[b*3-2],s2[b*3-1],s2[b*3],s2[c*3-2],s2[c*3-1],s2[c*3],s2[d*3-2],s2[d*3-1],s2[d*3]); printf("%c%c%c%c%c%c%c%c%c%c%c%c\n",s3[a*3-2],s3[a*3-1],s3[a*3],s3[b*3-2],s3[b*3-1],s3[b*3],s3[c*3-2],s3[c*3-1],s3[c*3],s3[d*3-2],s3[d*3-1],s3[d*3]); } return 0; }这是最开始写的纯暴力方法，看上去好不协调啊…+_+…
#include <algorithm> #include <iostream> #include <cstring> #include <cstdlib> #include <cstdio> #include <string> #include <vector> #include <cmath> #include <ctime> #include <queue> #include <stack> #include <set> #include <map> using namespace std; typedef long long LL; const int maxn= 100000 + 10; char s[3][35]= {" _ _ _ _ _ _ _ _ ", " | _| _||_||_ |_ ||_||_|| |", " ||_ _| | _||_| ||_| _||_|" }; int a[4]; int main() { while(~scanf("%d",&a[0])) { for(int i=1; i<4; i++) scanf("%d",&a[i]); for(int i=0; i<4; i++) if(!a[i]) a[i]+=10; for(int i=0; i<3; i++) { for(int j=0; j<4; j++) { int t=a[j]; printf("%c%c%c",s[i][t*3-2],s[i][t*3-1],s[i][t*3]); } puts(""); } } return 0; }O(∩_∩)O哈哈~这样看上去清爽多了吧。
还有一种比较屌的方法，是照着基神的代码敲的（真不知道基神脑袋怎么长的，就是跟常人思维不一样……好吧…%>_<%…其实不可否认的还是自己太愚笨…Orz），虽然本人目前没看懂，但还是贴上来把，求大神赐教^_^
#include <algorithm> #include <iostream> #include <cstring> #include <cstdlib> #include <cstdio> #include <string> #include <vector> #include <cmath> #include <ctime> #include <queue> #include <stack> #include <set> #include <map> using namespace std; typedef long long LL; const int maxn= 100000 + 10; char s[50][50]; int data[]={119,36,93,109,46,107,123,37,127,111}; int slove(int p,int x) { x=data[x]; if(x>>0&1)s[0][p+1]='_'; if(x>>1&1)s[1][p+0]='|'; if(x>>2&1)s[1][p+2]='|'; if(x>>3&1)s[1][p+1]='_'; if(x>>4&1)s[2][p+0]='|'; if(x>>5&1)s[2][p+2]='|'; if(x>>6&1)s[2][p+1]='_'; } int main() { int a[5]; while(~scanf("%d",&a[0])) { for(int i=1; i<4; i++) scanf("%d",&a[i]); memset(s,' ',sizeof(s)); for(int i=0;i<3;i++) s[i][12]=0; for(int i=0;i<4;i++) slove(i*3,a[i]); for(int i=0;i<3;i++) printf("%s\n",s[i]); } return 0; }
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时间： 2024-10-25 19:05:37