pat_1005

PAT A 1005 题解

最新推荐文章于 2024-07-19 00:01:03 发布

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本文提供了一道PAT A类竞赛题1005的解答方案，该题要求输入一个字符串并输出其各个数字字符之和的英文单词表达。代码中包含了完整的C++实现细节，并进行了特判处理。

源地址：http://www.patest.cn/contests/pat-a-practise/1005

模拟题，注意一下，ans可能为0，这个需要特殊判断，考虑全面就行了。

#include<stdio.h>
#include<iostream>
#include<string>
#include<string.h>
#include<algorithm>
#include<iomanip>
#include<vector>
#include<time.h>
#include<queue>
#include<stack>
#include<iterator>
#include<math.h>
#include<stdlib.h>
#include<limits.h>
#include<set>
#include<map>
//#define ONLINE_JUDGE
#define eps 1e-3
#define INF 0x7fffffff
#define FOR(i,a) for((i)=0;i<(a);(i)++)
#define MEM(a) (memset((a),0,sizeof(a)))
#define sfs(a) scanf("%s",a)
#define sf(a) scanf("%d",&a)
#define sfI(a) scanf("%I64d",&a)
#define pf(a) printf("%d\n",a)
#define pfI(a) printf("%I64d\n",a)
#define pfs(a) printf("%s\n",a)
#define sfd(a,b) scanf("%d%d",&a,&b)
#define sft(a,b,c)scanf("%d%d%d",&a,&b,&c)
#define for1(i,a,b) for(int i=(a);i<b;i++)
#define for2(i,a,b) for(int i=(a);i<=b;i++)
#define for3(i,a,b)for(int i=(b);i>=a;i--)
#define MEM1(a) memset(a,0,sizeof(a))
#define MEM2(a) memset(a,-1,sizeof(a))
const double PI=acos(-1.0);
template<class T> T gcd(T a,T b){return b?gcd(b,a%b):a;}
template<class T> T lcm(T a,T b){return a/gcd(a,b)*b;}
template<class T> inline T Min(T a,T b){return a<b?a:b;}
template<class T> inline T Max(T a,T b){return a>b?a:b;}
using namespace std;
#define ll __int64
int n,m;
#define Mod 1000000007
#define N 510
#define M 1000100
const int size = 10010;
const int mod = 9901;
int vis[N];
char ch[15][15] = {"zero","one","two","three","four","five","six","seven","eight","nine"};
char s[N];
int main(){
#ifndef ONLINE_JUDGE
    freopen("in.txt","r",stdin);
//  freopen("out.txt","w",stdout);
#endif
    while(scanf("%s",s)!=EOF){
    	int len = strlen(s);
    	int ans = 0;
    	for(int i=0;i<len;i++)
    		ans += (s[i]-'0');
    	if(ans == 0){						//特判
    		printf("%s\n",ch[0]);
    		continue;
    	}
    	vector<int> v;
    	while(ans){
    		int h = ans%10;
    		v.push_back(h);
    		ans /= 10;
    	}
    	int size = (int)v.size();
    	printf("%s",ch[v[size-1]]);
    	for(int i=size-2;i>=0;i--)
    		printf(" %s",ch[v[i]]);
    	printf("\n");
    }
return 0;
}