PAT1044. Shopping in Mars (25)(二分)

题意：

给出一系列数字，要求输出子序列使其和等于m，如果没有等于m的子序列就输出和减去m的差最小的子序列

要点：

这题的时间要求是100ms，n的量级是10^5，O(n^2)是肯定不行了，但我一开始算了一下O(nlogn)觉得不大行，就一直在想O(n)的方法，实际上O(nlogn)=10^5*17=1.7*10^6差不多是行的。接下来二分就行了。

#include<iostream>
#include<string>
#include<vector>
#include<map>
#include<set>
#include<stack>
#include<sstream>
#include<functional>
#include<algorithm>
using namespace std;
const int INF = 0xfffff;
const int maxn = 100050;
int num[maxn],sum[maxn];
vector<pair<int, int>> res;
int n, m;

void binery(int i, int &j, int &tempMin) {
	int left = i, right = n;
	while (left < right) {
		int mid = (left + right) / 2;
		if (sum[mid] - sum[i - 1] < m) {
			left = mid+1;
		} else
			right = mid;
	}
	j = right;
	tempMin = sum[j] - sum[i - 1];
}

int main() {
	scanf("%d%d", &n,&m);
	for (int i = 1; i <= n; i++) {
		scanf("%d", &num[i]);
		sum[i] = sum[i - 1] + num[i];
	}
	int Min = sum[n];
	for (int i = 1; i <= n; i++) {
		int j, tempMin;
		binery(i, j, tempMin);
		//cout << j << " " << tempMin << endl;
		if (tempMin >= m) {
			if (Min > tempMin) {
				Min = tempMin;
				res.clear();
				res.push_back({ i,j });
			} else if (Min == tempMin) {
				res.push_back({ i,j });
			}
		}
	}
	for (auto x : res) {
		printf("%d-%d\n", x.first, x.second);
	}
	return 0;
}