pat_1005

最新推荐文章于 2024-07-19 00:01:03 发布

ThinkingLion

最新推荐文章于 2024-07-19 00:01:03 发布

阅读量337

点赞数

分类专栏： PAT 文章标签：模拟

本文链接：https://blog.csdn.net/ThinkingLion/article/details/44702949

版权

PAT 专栏收录该内容

13 篇文章

订阅专栏

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;
}