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1044. Shopping in Mars (25)

2015-08-09 10:13 495 查看

1044. Shopping in Mars (25)

时间限制100 ms 内存限制65536 kB 代码长度限制16000 B 判题程序Standard作者CHEN, Yue 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:1. 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). 2. Cut before 5 or after 6, and take off the diamonds from the position 4 to 6 (with values 5+4+6=15). 3. 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
题目N个数,看看是否有连续的几个数和是M的,是按照i顺序输出i-j; 否则找到一个min>M又最接近M的那个,输出连续几个和是min的,按照i顺序输出i-j;我用到的queue 先进先出队列

评测结果

时间结果得分题目语言用时(ms)内存(kB)用户
8月09日 10:02答案正确251044C++ (g++ 4.7.2)242612datrilla

测试点

测试点结果用时(ms)内存(kB)得分/满分
0答案正确129610/10
1答案正确13087/7
2答案正确105642/2
3答案正确2426122/2
4答案正确11802/2
5答案正确95642/2
#include<iostream>#include<queue>struct ShoppinginMars{  int sT, eN;  ShoppinginMars(int s, int e) :sT(s), eN(e){}};using namespace std;    int main(){  queue<ShoppinginMars>OQ;  int N,M,sum,index,min;  int *options;  ShoppinginMars s(0,0);  scanf("%d%d", &N, &M);  options = new int;  for (index = 0,sum=0,min=M; index < N; index++)  {    scanf("%d", &options[index]);    sum += options[index];    s.eN = index;    while (sum >= M&&s.sT<=s.eN)    {       if (M == min || min > sum)min = sum;      if (sum == M)      {        OQ.push(s);       }       sum -= options[s.sT];      s.sT++;    }    }  if (OQ.empty())  {    for (index = 0, sum = 0,s.sT=0; index < N; index++)    {       sum += options[index];      s.eN = index;      while (sum >= min&&s.sT <= s.eN)      {         if (sum == min)        {          OQ.push(s);        }        sum -= options[s.sT];        s.sT++;      }    }  }   while (!OQ.empty())   {    printf("%d-%d\n",OQ.front().sT +1,OQ.front().eN +1);    OQ.pop();   }   delete[]options;  system("pause");  return 0;} 
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