Read Phone Number

Description
Do you know how to read the phone numbers in English? Now let me tell you.

For example, In China, the phone numbers are 11 digits, like: 15012233444. Someone divides the numbers into 3-4-4 format, i.e. 150 1223 3444. While someone divides the numbers into 3-3-5 format, i.e. 150 122 33444. Different formats lead to different ways to read these numbers:

150 1223 3444 reads one five zero one double two three three triple four.

150 122 33444 reads one five zero one double two double three triple four.

Here comes the problem:

Given a list of phone numbers and the dividing formats, output the right ways to read these numbers.

Rules:

Single numbers just read them separately.

2 successive numbers use double.

3 successive numbers use triple.

4 successive numbers use quadruple.

5 successive numbers use quintuple.

6 successive numbers use sextuple.

7 successive numbers use septuple.

8 successive numbers use octuple.

9 successive numbers use nonuple.

10 successive numbers use decuple.

More than 10 successive numbers read them all separately.

Input
The first line of the input gives the number of test cases, T(1 ≤ T ≤ 100).

T test cases follow. Each line contains a phone number N(1 ≤ length of N ≤ 100) and the dividing format F, one or more positive integers separated by dashes (-), without leading zeros and whose sum always equals the number of digits in the phone number.

Output
For each test case, output one line containing “Case #x: y”, where x is the case number (starting from 1) and y is the reading sentence in English whose words are separated by a space.
Sample Input
3
15012233444 3-4-4
15012233444 3-3-5
12223 2-3
Sample Output
Case #1: one five zero one double two three three triple four
Case #2: one five zero one double two double three triple four
Case #3: one two double two three

意思就是将号码分块读，12 223分开读就是一个1 和两个2 一个3
15555 3-2 就是分成155 55 读就是一个1 两个5和两个5

#include <iostream>#include <cstdio>#include <cstring>using namespace std;int main(){    int n;    cin>>n;    char num[9][10]= {"double","triple","quadruple","quintuple","sextuple","septuple","octuple","nonuple","decuple"};    char aa[10][6]= {"zero","one","two","three","four","five","six","seven","eight","nine"};    for(int k=1; k<=n; k++)    {        char a[101];//号码数组        scanf("%s",a);        int b[200]= {0};//每个分区的长度        char c[205];//3-4-4字符串的读入        scanf("%s",c);        int l=0;        for(int i=0; i<strlen(c); i++)        {            if(c[i]>='0'&&c[i]<='9')                b[l]=b[l]*10+c[i]-'0';            else                l++;        }        printf("Case #%d:",k);        int s=0;        for(int i=0; i<=l; i++)//操作输出        {            int sum=1;            if(b[i]==1)//一个时特例，直接输出                cout<<" "<<aa[a[s]-'0'];            else                for(int j=s; j<s+b[i]; j++)                {                    if(j!=s)                    {                        if(a[j]==a[j-1])                        {                            sum++;                        }                        else                        {                            if(sum>10)//超过十个也是一个个输出                            {                                for(int q=0;q<sum;q++)                                    cout<<" "<<aa[a[j-1]-'0'];                            }                            else if(sum>=2)//两个到十个得用double等输出                                cout<<" "<<num[sum-2]<<" "<<aa[a[j-1]-'0'];                            else//一个的输出                                cout<<" "<<aa[a[j-1]-'0'];                            sum=1;                        }                        if(j==s+b[i]-1)//每个分区的最后一个的输出                        {                            if(sum>10)                            {                                for(int q=0;q<sum;q++)                                    cout<<" "<<aa[a[j]-'0'];                            }                            else if(sum>=2)                                cout<<" "<<num[sum-2]<<" "<<aa[a[j]-'0'];                            else                                cout<<" "<<aa[a[j]-'0'];                            sum=1;                        }                    }                }            s+=b[i];        }        cout<<endl;    }    return 0;}