NOJ coder的音乐烦恼--字符串处理

http://acm.nit.net.cn/showproblem.jsp?pid=1366

#include<iostream>
#include<algorithm>
#include<string>
#include<cmath>

using namespace std;

int const ic_max = 5001;

struct stuNode
{
    int iPriority;//优先级
    string strSong;//歌曲名
};

void vInputData(int iN,stuNode stuSong[]);
void vGetData(string strT,stuNode& stuSong);
bool bCmp(const stuNode& stuA,const stuNode& stuB);
void vPrintAns(int iN,stuNode stuSong[]);

int main()
{
    int iNum;
    stuNode stuSong[ic_max];

    while(cin >> iNum)
    {
        vInputData(iNum,stuSong);
        sort(stuSong+1,stuSong+1+iNum,bCmp);
        vPrintAns(iNum,stuSong);
    }

    return 0;
}

void vInputData(int iN,stuNode stuSong[])
{
    string strTemp;

    getchar();
    for(int i=1; i<=iN; i++)
    {
        getline(cin,strTemp);//默认/n
        vGetData(strTemp,stuSong[i]);
    }
}

void vGetData(string strT,stuNode& stuSong)
{
    int iPrio;
    int iBlank;//标记最后一个空格的位置
    string strPrio;//获取优先级的字符串形式

    iPrio = 0;
    //获取最后一个空格的位置
    for(int i=strT.size()-1; i>=0; i--)
    {
        if(strT[i] == ' ')
        {
            iBlank = i;
            break;
        }
    }
    //截取相应子段
    stuSong.strSong = strT.substr(0,iBlank);
    strPrio = strT.substr(iBlank+1,strT.size()-iBlank);
    //将字符串转换为数字
    for(int i=strPrio.size()-1; i>=0; i--)
    {
        if(strPrio[i] == '-')
        {
            iPrio *= -1;
        }
        iPrio += (strPrio[i]-'0') * (int)pow(10.0,1.0*(strPrio.size()-1-i));
    }
    stuSong.iPriority  = iPrio;
}

bool bCmp(const stuNode& stuA,const stuNode& stuB)
{
    return stuA.iPriority < stuB.iPriority;
}

void vPrintAns(int iN,stuNode stuSong[])
{
    int iTotal;

    iTotal = iN;
    //获取优先级<=0的个数
    for(int i=1; i<=iN; i++)
    {
        if(stuSong[i].iPriority <= 0)
            iTotal --;
        else
            break;
    }
    cout << iTotal << endl;
    for(int i=iN-iTotal+1; i<=iN; i++)
    {
        cout << stuSong[i].strSong << endl;
    }
}