PAT 甲级 1044 Shopping in Mars (25 分)-蒲公英云

PAT 甲级 1044 Shopping in Mars (25 分)

野性酷女 2024-04-17 18:05 128阅读 0赞

1044 Shopping in Mars (25 分)

Shopping in Mars is quite a different experience. The Mars people pay by chained diamonds. Each diamond has a value (in Mars dollars M$). When making the payment, the chain can be cut at any position for only once and some of the diamonds are taken off the chain one by one. Once a diamond is off the chain, it cannot be taken back. For example, if we have a chain of 8 diamonds with values M$3, 2, 1, 5, 4, 6, 8, 7, and we must pay M$15. We may have 3 options:

Cut the chain between 4 and 6, and take off the diamonds from the position 1 to 5 (with values 3+2+1+5+4=15).
Cut before 5 or after 6, and take off the diamonds from the position 4 to 6 (with values 5+4+6=15).
Cut before 8, and take off the diamonds from the position 7 to 8 (with values 8+7=15).

Now given the chain of diamond values and the amount that a customer has to pay, you are supposed to list all the paying options for the customer.

If it is impossible to pay the exact amount, you must suggest solutions with minimum lost.

Input Specification:

Each input file contains one test case. For each case, the first line contains 2 numbers: N (≤105), the total number of diamonds on the chain, and M (≤108), the amount that the customer has to pay. Then the next line contains N positive numbers D1⋯DN (Di≤103 for all i=1,⋯,N) which are the values of the diamonds. All the numbers in a line are separated by a space.

Output Specification:

For each test case, print i-j in a line for each pair of i ≤ j such that Di + … + Dj = M. Note that if there are more than one solution, all the solutions must be printed in increasing order of i.

If there is no solution, output i-j for pairs of i ≤ j such that Di + … + Dj >M with (Di + … + Dj −M) minimized. Again all the solutions must be printed in increasing order of i.

It is guaranteed that the total value of diamonds is sufficient to pay the given amount.

Sample Input 1:

16 15
3 2 1 5 4 6 8 7 16 10 15 11 9 12 14 13

Sample Output 1:

1-5
4-6
7-8
11-11

Sample Input 2:

5 13
2 4 5 7 9

Sample Output 2:

2-4
4-5

思路：

1.如果sum[i](只从开始到第i个累加和)<m,那么它的字数列也会小于，则不用考虑。
2.如果sum[i](只从开始到第i个累加和)>m,那么我们循环减掉前面的，直到它等于m，你就找到答案了。
3.如果找不到，可在上述第二步中记录一个最小值。

#include<iostream>
#include<cmath>
using namespace std;
int sum[100005]={0};
int main()
{
    int n,m,i,j,d,t;
    bool isfind=false;
    d=100000005;
    j=0;
    scanf("%d %d",&n,&m);
    for(i=1;i<=n;i++)
    {
        scanf("%d",&d);
        sum[i]=sum[i-1]+d;
        while(sum[i]-sum[j]>m)
        {
            d=min(sum[i]-sum[j]-m,d);
            j++;
        }
        if(sum[i]-sum[j]==m)
        {
            printf("%d-%d\n",j+1,i);
            isfind=true;
        }
    }
    if(isfind==false)
    {
        j=0;
        for(i=1;i<=n;i++)
        {
            while(sum[i]-sum[j]-m>d)
            j++;
            if(sum[i]-sum[j]-m==d)
            {
                printf("%d-%d\n",j+1,i);
            }
        }
    }
    return 0;
}

大佬的代码：

思路：

解题思路：双指针法，一次遍历，p为左指针，q为右指针，取p=1时，q就遍历到p刚好是p-q的对应的数组的值的和大于等于目标结果，同时设一个临时结果，若此时的结果小于临时结果，且大于等于目标结果就清空目标数组，若等于，则直接存储此时的ij位置，然后p向前遍历，q也相应的前移以再次满足大于等于目标结果的。

#include <iostream>
#include <vector>
using namespace std;
int main(){
    int n, target, p = 0, q = 1, ans = 0, minans = 0x7fffffff;
    scanf("%d %d", &n, &target);
    vector<int>money(n + 1), result;
    money[0] = 0;
    for(int i = 1; i <= n ; ++ i)
        scanf("%d", &money[i]);
    for(; p <= n; ++ p){
        ans -= money[p];
        while(q <= n && ans < target)
            ans += money[q++];
        if(ans >= target && ans < minans){
            minans = ans;
            result.clear();
            result.push_back(p + 1);
            result.push_back(q - 1);
        }
        else if(ans == minans){
            result.push_back(p + 1);
            result.push_back(q - 1);
        }
    }
    for(int i = 0; i < result.size(); i += 2)
        printf("%d-%d\n",result[i], result[i + 1]);
}