1044. Shopping in Mars (25)

分析：

　　考察二分，简单模拟会超时，优化后时间正好，但二分速度快些，注意以下几点：

　　　　(1)：如果一个序列D₁ ... D_n，如果我们计算D_i到D_j的和， 那么我们可以计算D₁到D_j的和sum1，D₁到D_i的和sum2， 然后结果就是sum1 - sum2；

　　　　(2): 那么我们二分则要搜索的就是m + sum[i]的值。

#include <iostream>

#include <stdio.h>

#include <algorithm>

#include <cstring>

#include <string>

#include <vector>

#include <cctype>

#include <iomanip>

#include <cmath>

#include <map>



using namespace std;



const int Max_Int = 0x7fffffff;

const int Max_required = 100005;



int sum[Max_required];

int value[Max_required];



struct Node

{

    int start;

    int end;

};

vector<Node> V_node;



int binary_find(int target, int n)

{

    int low = 0, high = n;



    while (low <= high)

    {

        int mid = (low + high) >> 1;

        if (sum[mid] == target)

            return mid;

        else if (sum[mid] < target)

            low = mid + 1;

        else high = mid - 1;

    }

    return low;

}



int main()

{

    int n, m;

    while (scanf("%d%d", &n, &m) != EOF)

    {

        for (int i = 1; i <= n; i++)

        {

            scanf("%d", &value[i]);

            sum[i] = sum[i - 1] + value[i];

        }



        int Min = Max_Int;

        for (int i = 0; i <= n; i++)

        {

            int target = m + sum[i];

            int res = binary_find(target, n);

            //printf("res = %d\n", res);

            if (sum[res] - sum[i] - m >= 0 && sum[res] - sum[i] - m <= Min)

            {

                if (sum[res] - sum[i] - m < Min)

                {

                    V_node.clear();

                    Min = sum[res] - sum[i] - m;

                    Node node;

                    node.start = i + 1;

                    node.end = res;

                    V_node.push_back(node);

                }

                else if (sum[res] - sum[i] - m == Min)

                {

                    Node node;

                    node.start = i + 1;

                    node.end = res;

                    V_node.push_back(node);

                }

                

            }

        }



        for (int i = 0; i < V_node.size(); i++)

            printf("%d-%d\n", V_node[i].start, V_node[i].end);

    }

    return 0;

}

View Code