ZOJ3487-Ordinal Numbers

Input

There are multiple test cases. The first line of input is an integer T ≈ 1000 indicating the number of test cases.

Each test case consists of a cardinal number 0 ≤ n < 1,000,000,000.

Output

For each test case, output the corresponding ordinal number.

Sample Input

512341024
Sample Output
1st2nd3rd4th1024th
References
http://en.wikipedia.org/wiki/Names_of_numbers_in_Englishhttp://en.wikipedia.org/wiki/Ordinal_number_(linguistics)
Author: WU, ZejunContest: The 8th Zhejiang PRovincial Collegiate Programming Contest
题意：将阿拉伯数字后缀上序数词词尾。
#include <iostream>#include <cstdio>#include <string>#include <cstring>#include <algorithm>#include <cmath>#include <queue>#include <vector>#include <set>#include <stack>#include <map>#include <climits>using namespace std;#define LL long longconst int INF=0x3f3f3f3f;int main(){    int t;    scanf("%d",&t);    while(t--)    {        char ch[100];        scanf("%s",ch);        int len=strlen(ch);        if(len==1)        {            if(ch[0]=='1') printf("%sst/n",ch);            else if(ch[0]=='2') printf("%snd/n",ch);            else if(ch[0]=='3') printf("%srd/n",ch);            else printf("%sth/n",ch);        }        else        {            if(ch[len-2]=='1') printf("%sth/n",ch);            else            {                if(ch[len-1]=='1') printf("%sst/n",ch);                else if(ch[len-1]=='2') printf("%snd/n",ch);                else if(ch[len-1]=='3') printf("%srd/n",ch);                else printf("%sth/n",ch);            }        }    }    return 0;}