PAT Advanced 1005

题目如下:

1005. Spell It Right (20)

时间限制

400 ms

内存限制

65536 kB

代码长度限制

16000 B

判题程序

Standard

作者

CHEN, Yue

Given a non-negative integer N, your task is to compute the sum of all the digits of N, and output every digit of the sum in English.

Input Specification:

Each input file contains one test case. Each case occupies one line which contains an N (<= 10100).

Output Specification:

For each test case, output in one line the digits of the sum in English words. There must be one space between two consecutive words, but no extra space at the end of a line.

Sample Input:

12345

Sample Output:

one five

代码如下:

#include<iostream>
#include<string>
#include<sstream>
#include<vector>
using namespace std;
int sumdigits(string data);
template<class out, class in>
out convert(in a);
int main()
{
    string data;
    string digit[] = {"zero","one","two","three","four","five","six","seven","eight","nine"};
    string temp;
    while(cin>>data)
    {
        temp = convert<string>(sumdigits(data));
        for(int i = 0; i < temp.length(); i++)
        {
            cout<<digit[convert<int>(temp[i])];
            if (i != temp.length() - 1)
                cout<<" ";
        }

        cout<<endl;
    }
    return 0;
}

int sumdigits(string data)
{
    if ("" == data)
        return 0;
    int sum = 0;
    sum += convert<int>(data[data.length()-1]);
    sum += sumdigits(data.substr(0, data.length() - 1));
    return sum;

}
template<class out, class in>
out convert(in a)
{
    stringstream temp;
    temp<<a;
    out b;
    temp>>b;
    return b;
}
时间: 2024-11-05 06:25:55

PAT Advanced 1005的相关文章

PAT Advanced 1005 Spell It Right

Given a non-negative integer N, your task is to compute the sum of all the digits of N, and output every digit of the sum in English. Input Specification: Each input file contains one test case. Each case occupies one line which contains an N (≤). Ou

Pat(Advanced Level)Practice--1043(Is It a Binary Search Tree)

Pat1043代码 题目描述: A Binary Search Tree (BST) is recursively defined as a binary tree which has the following properties: The left subtree of a node contains only nodes with keys less than the node's key. The right subtree of a node contains only nodes

Pat(Advanced Level)Practice--1076(Forwards on Weibo)

Pat1076代码 题目描述: Weibo is known as the Chinese version of Twitter. One user on Weibo may have many followers, and may follow many other users as well. Hence a social network is formed with followers relations. When a user makes a post on Weibo, all hi

PAT甲级1005 Spell It Right

题目:PAT甲级 1005 题解:水题.看到题目的第一时间就在想一位一位的mod,最后一加一转换就完事了.结果看到了N最大为10的100的次方,吓得我赶紧放弃这个想法... 发现碰到这种情况用字符串十分好用,这道题应该考察的就是这一点.大致思路就是把数字的每一位放到字符串中,然后通过ASCII码得到每一位的相加结果num,然后把num一位一位的放到stack中,使用stack是因为它先进先出的特性,最后输出就行了. 代码: 1 #include<cstdio> 2 #include<qu

Pat(Advanced Level)Practice--1044(Shopping in Mars)

Pat1044代码 题目描述: Shopping in Mars is quite a different experience. The Mars people pay by chained diamonds. Each diamond has a value (in Mars dollars M$). When making the payment, the chain can be cut at any position for only once and some of the diam

PAT (Advanced Level) 1093. Count PAT&#39;s (25)

预处理每个位置之前有多少个P,每个位置之后有多少个T. 对于每个A,贡献的答案是这个A之前的P个数*这个A之后T个数. #include<cstdio> #include<cstring> long long MOD=1e9+7; const int maxn=1e5+10; long long dp1[maxn],dp2[maxn]; char s[maxn]; int main() { scanf("%s",s); memset(dp1,0,sizeof d

PAT (Advanced Level) 1055. The World&#39;s Richest (25)

排序.随便加点优化就能过. #include<iostream> #include<cstring> #include<cmath> #include<algorithm> #include<cstdio> #include<map> #include<queue> #include<string> #include<stack> #include<vector> using names

Pat(Advanced Level)Practice--1018(Public Bike Management)

Pat1018代码 题目描述: There is a public bike service in Hangzhou City which provides great convenience to the tourists from all over the world. One may rent a bike at any station and return it to any other stations in the city. The Public Bike Management C

Pat(Advanced Level)Practice--1016(Phone Bills)

Pat1016代码 题目描述: A long-distance telephone company charges its customers by the following rules: Making a long-distance call costs a certain amount per minute, depending on the time of day when the call is made. When a customer starts connecting a lon