PAT A1009-1012

A 1009 Product of Polynomials (25 point(s))

　　读懂题意就行。

#include <cstdio>
#include <iostream>
#include <vector>
#include <algorithm>

using namespace std;
typedef struct NODE
{
    int exp;
    double coe;
    NODE(){exp = 0; coe = 0;}
    NODE(int e, double v):exp(e),coe(v){}
}node;
vector<node> poly1, poly2;
double rstPoly[2010]={0};
int main()
{
    int N, tmpExp;
    double tmpCoe;
    cin >> N;
    for(int i = 0; i < N; ++ i)
    {
        cin >> tmpExp >> tmpCoe;
        poly1.push_back(NODE(tmpExp, tmpCoe));
    }
    cin >> N;
    for(int i = 0; i < N; ++ i)
    {
        cin >> tmpExp >> tmpCoe;
        poly2.push_back(NODE(tmpExp, tmpCoe));
    }
    for(int i = 0; i < poly1.size(); ++i)
        for(int j = 0; j < poly2.size(); ++ j)
            rstPoly[poly1[i].exp+poly2[j].exp] += poly1[i].coe*poly2[j].coe;
    int cnt = 0;
    for(int i = 0; i < 2010; ++ i)
        if(rstPoly[i] != 0)
            cnt++;
    cout << cnt;
    for(int i = 2000; i >= 0; -- i)
        if(rstPoly[i] != 0)
        {
            printf(" %d %.1f", i, rstPoly[i]);
        }
    return 0;
}

A 1010 Radix (25 point(s))

　　这道题目相当有趣，坑的莫名其妙，哈哈哈。

　　注意：1.数的范围，10位数字，int 不够用

　　　　　2.超时，因而不能从小到大递增取进制数

　　　　   3.二分法，最大进制和最小进制的确定

　　　　   4.超限，在找进制数的过程中，过大的进制数会发生超限，注意这种情况

　　　　　5.待确定进制数只有一位，此时需要注意取最小进制

#include <cstdio>
#include <iostream>
#include <cstring>
#include <cstdlib>
#include <vector>
#include <string>
#include <algorithm>

using namespace std;
long long toNum(char c)
{
    return isdigit(c) ? c-'0' : c-'a'+10;
}
long long strRadixToNum(string tmpStr, long long radix)
{
    long long tmpNum = 0, ant = 1;
    for(int i = tmpStr.size()-1; i >= 0; -- i)
    {
        tmpNum += toNum(tmpStr[i])*ant;
        if(tmpNum < 0 || ant < 0)
            return -1;
        ant *= radix;
    }
    return tmpNum;
}
long long minRadix(string tmpStr)
{
    char tmpChar = 0;
    for(int i = 0; i < tmpStr.size(); ++ i)
    {
        if(tmpStr[i] > tmpChar)
            tmpChar = tmpStr[i];
    }
    return toNum(tmpChar) + 1;
}
int main()
{
    long long tag, radix, tmpNum, tmpTarget, lowRadix, highRadix;
    string N1, N2;
    cin >> N1 >> N2 >> tag >> radix;
    //算出进制已经确定的数作为目标值
    //并且 找出未确定进制的数的最小进制
    if(tag == 2)
        swap(N1, N2);
    highRadix = tmpTarget = strRadixToNum(N1, radix);
    lowRadix = max((long long)2 , minRadix(N2));
    while(lowRadix <= highRadix)
    {
        long long midRadix = (lowRadix + highRadix)/2;
        long long tmpNum1 = strRadixToNum(N2, midRadix);
        if(tmpNum1 >= tmpTarget ||tmpNum1 == -1)
             highRadix = midRadix - 1;
        else
            lowRadix = midRadix + 1;
    }
    if(strRadixToNum(N2, lowRadix) == tmpTarget)
        cout << lowRadix;
    else
        cout << "Impossible";    
    return 0;
}

 

A 1011 World Cup Betting (20 point(s))

#include <cstdio>
#include <iostream>
#include <cstring>
#include <cstdlib>
#include <vector>
#include <algorithm>

using namespace std;
int main()
{
    double sum = 0.65, tmpW, tmpT, tmpL;
    for(int i = 0; i < 3; ++ i)
    {
        cin >> tmpW >> tmpT >> tmpL;
        if(tmpW > max(tmpT, tmpL))
        {
            cout << "W ";
            sum *= tmpW;
        }
        else if(tmpT > max(tmpW, tmpL))
        {
            cout << "T ";
            sum *= tmpT;
        }
        else
        {
            cout << "L ";
            sum *= tmpL;;
        }
    }
    printf("%.2f", (sum-1)*2);
    return 0;
}

A 1012 The Best Rank (25 point(s))

　　注意：1.排名的问题，相同成绩同一排名

　　　　　2.最佳排名相同，按所给优先级

#include <cstdio>
#include <iostream>
#include <cstring>
#include <cstdlib>
#include <vector>
#include <algorithm>
#include <set>
#include <unordered_map>
#include <functional>

using namespace std;
typedef struct NODE
{
    int id, math, cCode, aver, english;
}node;
vector<int> cCodeVec, mathVec, engVec, averVec;
unordered_map<int, node> nodeMap;
unordered_map<int, int> nodeTableMap;
void getBestRank(int u)
{
    int type = 0, rank, cnt;
    char typeStr[5]="ACME";
    node tmpNode = nodeMap[u];
    cnt = 1;
    for(auto it = averVec.begin(); it != averVec.end(); ++it, ++cnt)
        if(*it == tmpNode.aver)
        {
            type = 0; rank = cnt;break;
        }
    cnt = 1;
    for(auto it = cCodeVec.begin(); it != cCodeVec.end(); ++it, ++cnt)
        if(*it == tmpNode.cCode && cnt < rank)
        {
            type = 1; rank = cnt;break;
        }
    cnt = 1;
    for(auto it = mathVec.begin(); it != mathVec.end(); ++it, ++cnt)
        if(*it == tmpNode.math && cnt < rank)
        {
            type = 2; rank = cnt;break;
        }
    cnt = 1;
    for(auto it = engVec.begin(); it != engVec.end(); ++it, ++cnt)
        if(*it == tmpNode.english && cnt < rank)
        {
            type = 3; rank = cnt;break;
        }
    printf("%d %c\n", rank, typeStr[type]);
}
int main()
{
    int N, M;
    node tmpNode;
    cin >> N >> M;
    for(int i = 0; i < N; ++i)
    {
        cin >> tmpNode.id >> tmpNode.cCode >> tmpNode.math >> tmpNode.english;
        tmpNode.aver = (tmpNode.cCode + tmpNode.math + tmpNode.english+1.5)/3;
        cCodeVec.push_back(tmpNode.cCode); mathVec.push_back(tmpNode.math);
        engVec.push_back(tmpNode.english); averVec.push_back(tmpNode.aver);
        nodeMap[tmpNode.id] = tmpNode;nodeTableMap[tmpNode.id] = 1;
    }
    sort(averVec.begin(), averVec.end(), greater<int>());
    sort(cCodeVec.begin(), cCodeVec.end(), greater<int>());
    sort(mathVec.begin(), mathVec.end(), greater<int>());
    sort(engVec.begin(), engVec.end(), greater<int>());
    for(int i = 0; i < M; ++ i)
    {
        cin >> tmpNode.id;
        if(nodeTableMap[tmpNode.id] > 0)
            getBestRank(tmpNode.id);
        else
            cout << "N/A" << endl;
    }
    return 0;
}