PAT Advanced Level 1044

1044 Shopping in Mars (25)（25 分）

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).

1. Cut before 5 or after 6, and take off the diamonds from the position 4 to 6 (with values 5+4+6=15).
2. 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 (<=10^5^), the total number of diamonds on the chain, and M (<=10^8^), the amount that the customer has to pay. Then the next line contains N positive numbers D~1~ ... D~N~ (D~i~<=10^3^ 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 D~i~ + ... + D~j~ = 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 D~i~ + ... + D~j~ > M with (D~i~ + ... + D~j~ - 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

/**********************
author: yomi
date: 18.6.10
ps: sum 递增  ---->  完全靠思路撑起一片天的模板题 
本题的巧妙之处就在于已知数列并不递增，让人想不到二分 
但是题目的条件Di+...+Dj>M 却给了暗示，仔细思考， 不难发现，序列和是递增的
那就可以用二分了。
1 2 3  4  5  6  7   8   9
3 2 1  5  4  6  8   7   16  10 15 11 9 12 14 13
3 5 6 11 15  21 29  36  52 
像这样。
那么i~j                                 j      i-1
下标4~6的序列和就是5+4+6 == 15， 也就是sum[6]-sum[3]
不难想象，进行二分的序列肯定是sum[]
通过上面的分析，可以知道m == sum[j]-sum[i-1]
所以 sum[j] == m+sum[i-1], 我们需要让bs()返回的就是j
所以需要查找的数就是m+sum[i-1].
接下来找到>=m的最小数nearm
然后再遍历一次，找到所有满足==nearm的i~j的sum
**********************/
#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstring>
using namespace std;
long long int sum[100010];


int bs(long long int l, long long int r, long long int s)
{
    int mid;
    while(l < r){
        mid = (l+r)/2;
        if(sum[mid] >= s){
            r = mid;
        }
        else{
            l = mid+1;
        }
    }
    return l;
}
int main()
{
    long long int n, m, nearm = 0, diff = 100000010;
    memset(sum, 0, sizeof(sum));
    scanf("%lld%lld", &n, &m);
    for(long long int i=1; i<=n; i++){
        scanf("%lld", &sum[i]);
        sum[i] += sum[i-1]; // sum[] 求和
    }
    for(long long int i=1; i<=n; i++){
        long long int j = bs(i, n+1, m+sum[i-1]);///二分下界是i而不是i+1，因为可能有一个数恰好为m
        long long int tmp = sum[j]-sum[i-1];
        if(tmp - m < diff && tmp >= m){
            diff = tmp-m;
            nearm = tmp;
        }
    }
    for(long long int i=1; i<=n; i++){
        long long int j = bs(i, n+1, sum[i-1]+nearm);
        if(sum[j]-sum[i-1] == nearm){
            printf("%lld-%lld
", i, j);
        }
    }
    return 0;
}

/**
16 15
3 2 1 5 4 6 8 7 16 10 15 11 9 12 14 13

1-5
4-6
7-8
11-11
**/