首页 > ACM题库 > HDU-杭电 > hdu 2329 Money Money Money, Must Be Funny待解决[解题报告]C++
2014
01-05

hdu 2329 Money Money Money, Must Be Funny待解决[解题报告]C++

Money Money Money, Must Be Funny

问题描述 :

* Shopkeeper: “100 crowns and 80 hellers, please.”
* Customer: “Here you are.” (handing a 200-crown banknote)
* Shopkeeper: “Would you have 80 hellers?” (passing back a 100-crown note)
* Customer: “No, but here you are.” (adding a 1-crown coin)
* Shopkeeper: “Sorry, but have only this.” (showing a little shiny 50-heller coin)
* Customer: “So I need to give you another 30 heller. But I have only 40.”
* Shopkeeper: “That’s ok, here are the remaining 10.”

Have you ever experienced similar situations? Paying a precise amount can sometimes be difficult, if the set of available coins and banknotes (“tenders”) is limited. The situation above was finally solved: The customer paid 200+1 crowns, got 100+0.50 back, paid another 0.20+0.20, and finally got 0.10 back. This means, 7 tenders had to be exchanged. Sometimes, it may be even more complicated. Your task is to write a program that solves situations like this.

输入:

Input contains several tasks to be solved. Each task begins with a line containing one nonnegative number: the amount to be paid. Then there is a list of tenders possessed by the customer (the one who pays). Each line in the list contains the tender nominal value (nonnegative number), one space, number of tenders of that value (non-negative integer), and the
lowercase letter “x”. The list is terminated by a line containing number “-1”.

After the first list, there is a list of tenders possessed by the shopkeeper (the one who gets paid). The second list has the exactly same format as the first one.

Then the next task begins. The last task is followed by one more line containing “-1”.

You may assume that each list will contain at most 100 lines, and nobody will have more than 10 000 units and/or 500 tenders. Nominal values can be arbitrary, they do not need to follow any existing scheme valid in known countries. All numbers that allow non-integer values will be given either as integers or as decimal numbers with one or two digits after the decimal point.

输出:

Input contains several tasks to be solved. Each task begins with a line containing one nonnegative number: the amount to be paid. Then there is a list of tenders possessed by the customer (the one who pays). Each line in the list contains the tender nominal value (nonnegative number), one space, number of tenders of that value (non-negative integer), and the
lowercase letter “x”. The list is terminated by a line containing number “-1”.

After the first list, there is a list of tenders possessed by the shopkeeper (the one who gets paid). The second list has the exactly same format as the first one.

Then the next task begins. The last task is followed by one more line containing “-1”.

You may assume that each list will contain at most 100 lines, and nobody will have more than 10 000 units and/or 500 tenders. Nominal values can be arbitrary, they do not need to follow any existing scheme valid in known countries. All numbers that allow non-integer values will be given either as integers or as decimal numbers with one or two digits after the decimal point.

样例输入:

100.80
500 1x
200 3x
1.00 10x
0.20 2x
-1
500 10x
200 12x
100 8x
0.10 1x
0.20 0x
0.50 100x
20 2x
-1
200
10 19x
-1
200 1x
-1
-1

样例输出:

7 tenders must be exchanged.
The payment is impossible.


  1. #include <cstdio>
    #include <algorithm>

    struct LWPair{
    int l,w;
    };

    int main() {
    //freopen("input.txt","r",stdin);
    const int MAXSIZE=5000, MAXVAL=10000;
    LWPair sticks[MAXSIZE];
    int store[MAXSIZE];
    int ncase, nstick, length,width, tmp, time, i,j;
    if(scanf("%d",&ncase)!=1) return -1;
    while(ncase– && scanf("%d",&nstick)==1) {
    for(i=0;i<nstick;++i) scanf("%d%d",&sticks .l,&sticks .w);
    std::sort(sticks,sticks+nstick,[](const LWPair &lhs, const LWPair &rhs) { return lhs.l>rhs.l || lhs.l==rhs.l && lhs.w>rhs.w; });
    for(time=-1,i=0;i<nstick;++i) {
    tmp=sticks .w;
    for(j=time;j>=0 && store >=tmp;–j) ; // search from right to left
    if(j==time) { store[++time]=tmp; }
    else { store[j+1]=tmp; }
    }
    printf("%dn",time+1);
    }
    return 0;
    }

  2. a是根先忽略掉,递归子树。剩下前缀bejkcfghid和后缀jkebfghicd,分拆的原则的是每个子树前缀和后缀的节点个数是一样的,根节点出现在前缀的第一个,后缀的最后一个。根节点b出现后缀的第四个位置,则第一部分为四个节点,前缀bejk,后缀jkeb,剩下的c出现在后缀的倒数第2个,就划分为cfghi和 fghic,第3部分就为c、c