1044. Shopping in Mars

Shopping in Mars is quite a different experience. The Mars people pay by chained diamonds. Each diamond have a value (in Mars dollars m$). When making the payment, the chain can is cut at any position for only once and some of the diamonds is taken off the CHA In one by one. Once a diamond is off the chain, it cannot was taken back. For example, if we had a chain of 8 diamonds with the 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).
2. Cut before 5 or after 6, and take off the diamonds from the position 4 to 6 (with values 5+4+6=15).
3. 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 a customer have to pay, you is supposed to list all the paying O Ptions 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 (<=108), the amount that the customer have to pay. Then the next line contains N positive numbers d_{1&NBSP; Dn  (di<=103 for All I=1, ..., N) which is the values of the diamonds. All the numbers in a line is 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 + ... + dj = M. Note that if the Re is more than one solution, all the solutions must is printed in increasing order of I.}

If There is no solution, output "i-j" for pairs of I <= j such that D_{i + ... + dJ > M with (di + ... + D j-m) minimized. Again all the solutions must is printed in increasing order of I.}

It is guaranteed, the total value of diamonds was sufficient to pay the given amount.

Segment error ... I don't know, I'm not sure. The array has opened up very big ...

#include <stdio.h> #include <stdlib.h> #include <vector>using namespace std; #define N 10001typedef        struct{int start,end;        Long sum;    }node;int num[n];int N;long cost;int Main () {scanf ("%d%ld", &n,&cost);    int i,j;    Vector<node> equal;    Vector<node> min;    Long now=0,minsum=100000002;    Node temp;    for (i=1;i<=n;i++) scanf ("%d", num+i);         for (i=1;i<=n;i++) {now+=num[i];         J=i;                while (Now<cost&&j<n) {j + +;                NOW+=NUM[J];         } temp.start=i;         Temp.end=j;         Temp.sum=now;             if (now==cost) {equal.push_back (temp); } else if (temp.sum>cost) {if (temp.sum<minsum) {Minsum=te                 Mp.sum;                 Min.clear ();                 Min.push_back (temp); } else if (temp.sum==minsum) min.push_bACK (temp);         } now=0; } if (Equal.size () ==0) {for (I=0;i<min.size (); i++) {printf ("%d-%d\n", min[i].start,min             [I].end]; }} else {for (i=0;i<equal.size (); i++) {printf ("%d-%d\n", equal[i].start,equal[             I].end);    }} system ("Pause");    return 0; }

1044. Shopping in Mars