UVa Volume 1. Elementary Problem Solving :: String

Volume 1. Elementary Problem Solving :: String

string基础知识：http://hi.baidu.com/ybbaigo/blog/item/ce1490fb51e6a561034f5648.html

1. 401 - Palindromes

题目大意：判断一个字符串是不是回文串和镜像串

#include <iostream> #include <string> #include <cctype> using namespace std; int main() { char a[35]={'A',' ',' ',' ','3',' ',' ','H','I','L',' ','J','M',' ','O',' ',' ',' ','2','T','U','V','W','X','Y','5','1','S','E',' ','Z',' ',' ','8',' '}; string s; int l,i,flag1,flag2; while(cin>>s) { flag1=1;flag2=1;l=s.length(); for(i=0;i<l/2;i++)if(s[i]!=s[l-i-1]){flag1=0;break;} for(i=0;i<l;i++) { if(isalpha(s[l-i-1])){if(s[i]!=a[(int)s[l-i-1]-'A']){flag2=0;break;}} else if(s[i]!=a[(int)s[l-i-1]-'1'+26]){flag2=0;break;} } if(!flag1&&!flag2)cout<<s<<" -- is not a palindrome./n"; else if(flag1&&!flag2)cout<<s<<" -- is a regular palindrome./n"; else if(!flag1&&flag2)cout<<s<<" -- is a mirrored string./n"; else cout<<s<<" -- is a mirrored palindrome./n"; cout<<endl; } return 0; }

没什么技术含量，就是数组赋初值和字符转化为数组下标两点

2. 10010 - Where's Waldorf?

题目大意：在一个给定的字符矩阵中找到所给字符串，8个方向都行

#include <iostream> #include <string> #include <cctype> using namespace std; //ifstream fin("in.in"); string a[100],b; int flag,num,i,j,m,n,k,t,x,y,l; int main() { cin>>num; for(j=1;j<=num;j++) { cin>>m>>n; for(i=0;i<m;i++) { cin>>a[i]; for(k=0;k<a[i].length();k++)a[i][k]=toupper(a[i][k]); } cin>>k; for(i=0;i<k;i++) { cin>>b; l=b.length(); for(t=0;t<l;t++)b[t]=toupper(b[t]); for(y=0;y<n;y++) { flag=0; for(x=0;x<m;x++) { if(x+l-1<m) { flag=1; for(t=0;t<l;t++)if(a[x+t][y]!=b[t]){flag=0;break;} if(flag){cout<<x+1<<" "<<y+1<<endl;break;} } if(y+l-1<n) { flag=1; for(t=0;t<l;t++)if(a[x][y+t]!=b[t]){flag=0;break;} if(flag){cout<<x+1<<" "<<y+1<<endl;break;} } if(x-l+1>=0) { flag=1; for(t=0;t<l;t++)if(a[x-t][y]!=b[t]){flag=0;break;} if(flag){cout<<x+1<<" "<<y+1<<endl;break;} } if(y-l+1>=0) { flag=1; for(t=0;t<l;t++)if(a[x][y-t]!=b[t]){flag=0;break;} if(flag){cout<<x+1<<" "<<y+1<<endl;break;} } if(x+l-1<m&&y+l-1<n) { flag=1; for(t=0;t<l;t++)if(a[x+t][y+t]!=b[t]){flag=0;break;} if(flag){cout<<x+1<<" "<<y+1<<endl;break;} } if(x-l+1>=0&&y+l-1<n) { flag=1; for(t=0;t<l;t++)if(a[x-t][y+t]!=b[t]){flag=0;break;} if(flag){cout<<x+1<<" "<<y+1<<endl;break;} } if(x+l-1<m&&y-l+1>=0) { flag=1; for(t=0;t<l;t++)if(a[x+t][y-t]!=b[t]){flag=0;break;} if(flag){cout<<x+1<<" "<<y+1<<endl;break;} } if(x-l+1>=0&&y-l+1>=0) { flag=1; for(t=0;t<l;t++)if(a[x-t][y-t]!=b[t]){flag=0;break;} if(flag){cout<<x+1<<" "<<y+1<<endl;break;} } } if(flag)break; } } if(j!=num)cout<<endl; } return 0; }

首先是都要转化成大写字母，其次可以用方向数组来简化代码，不用写8个方向，只写一个就行了，再有flag=1;要写到if语句里面，否则会有很诡异的错误。