2012 winter training @HIT Day 4 解题报告

冬训第四天，练习贪心。春节之前最后一天的contest了，只有三道。

http://acm.hit.edu.cn/hoj/contest/view?id=100130

Problem A：Mixing Milk

hoj1003，以前刷过。就是贪心，挑最便宜的即可。

/*This Code is Submitted by acehypocrisy for Problem 4000095 at 2012-01-20 23:04:02*/
/*This Code is Submitted by acehypocrisy for Problem 1003 at 2011-09-19 09:24:59*/
#include <stdio.h>

int main(int argc, char *argv[])
{
 int total,n;
 scanf("%d %d",&total,&n);
 int farmers[5000][2];
 int q;
 for (q=0;q<n;q++)
         scanf("%d %d",&farmers[q][1],&farmers[q][2]);

 int i,j,tmp,flag;
 for (i=0;i<n-1;i++){
     flag=1;
     for (j=0;j<n-i-1;j++){
         if (farmers[j][1]>farmers[j+1][1]){
            tmp=farmers[j][1];
            farmers[j][1]=farmers[j+1][1];
            farmers[j+1][1]=tmp;

                        tmp=farmers[j][2];
            farmers[j][2]=farmers[j+1][2];
            farmers[j+1][2]=tmp;
            flag=0;
         }
         }
     if (1==flag)
        break;    
 }

 int sum=0,num=0,k=0;
 for (k=0;k<n;k++){
        if (num+farmers[k][2]<total){
                num+=farmers[k][2];
                sum+=farmers[k][1]*farmers[k][2];
        }else{
                sum+=(total-num)*farmers[k][1];
                break;
        }
 
 }


 printf("%d\n",sum);
 return 0;    
 }
 

Problem B：数列极差问题

每次删掉最小的两个，最后的数就是最大的；每次删掉最大的两个，最后的数就是最小的。

*This Code is Submitted by acehypocrisy for Problem 4000096 at 2012-01-21 13:34:06*/
#include <stdio.h>
#include <algorithm>

using namespace std;

bool compare (const int& a, const int& b){
    return a > b;
}

int main()
{
    int N;
    int sequence1[50000], sequence2[50000];
    while(scanf("%d", &N) == 1){
        if (N == 0)
            break;
        for (int i = 0; i < N; i++){
            scanf("%d", &sequence1[i]);
            sequence2[i] = sequence1[i];
        }
        while (N > 1){
            sort(sequence1, sequence1 + N);
            sort(sequence2, sequence2 + N, compare);
            sequence1[N - 2] = sequence1[N - 2] * sequence1[N - 1] + 1;
            sequence2[N - 2] = sequence2[N - 2] * sequence2[N - 1] + 1;
            N--;
        }
        printf("%d\n", sequence2[0] - sequence1[0]);
    }
    return 0;
}
 

Problem C：TEX Quotes

1101，以前也刷过……其实基本上应该没有太大的难度。

代码好像有点繁琐，其实一个字符一个字符的判断之后直接print出去就行，不必用还用string什么的。。当时可能没想到把。

cur指针记录的是从上一个 “ 到下一个 ” 之间的字符串。

/*This Code is Submitted by acehypocrisy for Problem 4000097 at 2012-01-20 23:02:56*/
/*This Code is Submitted by acehypocrisy for Problem 1101 at 2011-10-30 11:56:28*/
#include <cstdlib>
#include <iostream>
#include <string>

using namespace std;

int main(int argc, char *argv[])
{
    char temp[10000];
    string s="";
    int counter=0;
    while (cin.getline(temp,10000)){
          char* p=temp;
          char* cur=temp;
          while ((*p)!='\0'){
                if ((*p)=='\"'){
                   counter++;
                   (*p)='\0';
                   s+=cur;
                   if (counter%2==1){
                      s+="``";
                   }else{
                      s+="''";
                   }
                   p++;
                   cur=p;
                   continue;
                }
                p++;
          }
          s+=cur;
          cout<<s<<endl;
          s="";       
    }
    return 0;
}