PAT Advanced 1044 Shopping in Mars (25) [⼆分查找]

题目

Shopping in Mars is quite a diferent 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 of the chain one by one. Once a diamond is of the chain, it cannot be taken back. For example, if we have a chain of 8 diamonds with values M3, 2, 1, 5, 4, 6, 8, 7, and we must pay M15. We may have 3 options:

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

题目分析

已知一系列数S，已知一个整数N，一次从S中只能切分一次，切分之后可以从切分后的任意一边弹出M个数字，使得这M个数字之和>=M并且是最小值nearM，求出所有切分后可以弹出M个数字之和==nearM的情况，打印（"切分位置-最后一个弹出数字的位置"（从左往右弹出）或者"最后一个弹出数字的位置-切分位置"（从右往左弹出））

解题思路

思路 01（最优）

1. minans记录大于M的最小值，如果当前找到的tempnum>=M的最小值小于minans，清空之前记录的信息，因为找到了更小的情况
2. 将所有情况记录在vector中最后打印

思路 02

1. 二分查找到nearM>M的最小值（第一个大于M的数字的下标j，可能情况：s[j-1]-s[i-1] == M（nearM == M） 或者 s[j-1]-s[i-1]<M&&s[j]-s[i-1]>M（nearM == s[j]-s[i-1]））
2. 二分查找到s[j-1]-s[i-1]==nearM的所有情况

Code

Code 01

#include <iostream>
#include <vector>
using namespace std;
int N,M;
vector<int> sum;
vector<int> res;
void find(int i,int& j,int& tempnum) {
	int left =i,right=N;
	while(left<right) {
		int mid=(left+right)/2;
		if(sum[mid]-sum[i-1]>=M) {
			right = mid;
		} else {
			left = mid+1;
		}
	}
	j = right; //此时left==right
	tempnum=sum[j]-sum[i-1];
}
int main(int argc,char * argv[]) {
	// 1 输入
	scanf("%d%d",&N,&M);
//	int sum[N+1]= {0};
	sum.resize(N+1);
	for(int i=1; i<=N; i++) {
		scanf("%d",&sum[i]);
		sum[i]+=sum[i-1];
	}
	// 2 查找切分点
	int minans=sum[N]; //最小结果：若有等于M的minans=M，若没有等于M的取最小大于M的数作为minans的值
	for(int i=1; i<=N; i++) {
		int j,tempnum;
		find(i,j,tempnum); // 二分查找 找到>=M的第一个数 tempnum
		if(tempnum>minans) continue; //若tempnum比之前找到的最小值大，跳过不考虑 
		if(tempnum>=M) { //如果tempnum<=之前找到的最小值，并且>=M 
			if(tempnum<minans) { //如果小于 minans，替换 
				res.clear();
				minans=tempnum;
			}
			// 如果 tempnum==minans（1 tempnum等于当前大于M的最小值 2 tempnum==M） 
			res.push_back(i);
			res.push_back(j);
		}
	}
	for(int i=0; i<res.size(); i+=2) {
		printf("%d-%d
",res[i],res[i+1]);
	}
	return 0;
}

Code 02

#include <iostream>
using namespace std;
const int maxn = 100010;
const int inf = 0x7777777;
int sum[maxn],nearM=inf;
int upper_bound(int l,int r,int x) {
	int left =l,right=r,mid;
	while(left<right) {
		mid = left+((right-left)>>1);
		if(sum[mid]>x) right = mid;
		else left = mid+1;
	}
	return left;
}
int main(int argc,char * argv[]) {
	int n,m;
	scanf("%d%d",&n,&m);
	for(int i=1; i<=n; i++) {
		scanf("%d",&sum[i]);
		sum[i]+=sum[i-1];
	}
	// 找到第一个大于等于条件的最小序列和nearM
	for(int i=1; i<=n; i++) {
		int j = upper_bound(i,n+1,sum[i-1]+m); //找到第一个大于等于sum[i=1]+x的数
		if(sum[j-1]-sum[i-1]==m) { //查找成功
			nearM=m;
			break;
		} else if(j<=n&&sum[j]-sum[i-1]<nearM) {
			nearM=sum[j]-sum[i-1];
		}
	}
	// 找到满足nearM的第一个元素位置
	for(int i=1; i<=n; i++) {
		int j = upper_bound(i,n+1,sum[i-1]+nearM); //找到第一个大于nearM的位置 j 
		if(sum[j-1]-sum[i-1]==nearM) // j第一个大于，则j-1有可能等于 
			printf("%d-%d
",i,j-1);
	}
	return 0;
}