# PAT 解题报告 1044. Shopping in Mars (25)

### 1044. Shopping in Mars 题目描述：

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).
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 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.

More specifically:

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 sufficient to pay the given amount.

### 1044. Shopping in Mars 算法分析：

```#include <cstring>
#include <cstdio>
#include <cstdlib>
#include <cmath>

#include <algorithm>
#include <string>
#include <vector>
#define REP(i,n) for(int i=0;i<(n);++i)

using namespace std;

struct Info {
int sum;
int idx;
};

vector<Info> infos;
vector<int> diamonds;
int N, M;

void gao() {
if(0 == N) return;
infos[0].sum = 0;
int ix0 = -1;
while(infos[0].sum < M) {
ix0++;
infos[0].sum += diamonds[ix0];
}
infos[0].idx = ix0;

for(int i=1; i<N; ++i) {
infos[i].sum = infos[i-1].sum - diamonds[i-1];
int ix = infos[i-1].idx;
while(infos[i].sum < M && ix < N) {
ix++;
infos[i].sum += diamonds[ix];
}
infos[i].idx = ix;
if(infos[i].sum < M) {
infos[i].idx = -1;
break;
}
}
int mindiff = infos[0].sum - M;
REP(i, N) {
if(-1 == infos[i].idx) break;
if(infos[i].sum - M < mindiff) {
mindiff =  infos[i].sum - M;
}
if(0 == mindiff) break;
}

REP(i, N) {
if(-1 == infos[i].idx) break;
if(infos[i].sum - M == mindiff) {
printf("%d-%d\n", i+1, infos[i].idx+1);
}
}
return ;
}

int main(void) {
scanf("%d %d", &N, &M);
infos.resize(N);
diamonds.resize(N);
REP(i, N) scanf("%d", &diamonds[i]);
gao();
return 0;
}
```

### 1044. Shopping in Mars 注意点：

Written on January 18, 2014