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
", 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
", V_node[i].start, V_node[i].end);
    }
    return 0;
}