题目传送门
1 /*
2 题意：给出少了若干卡片后的总和，和原来所有卡片，问少了哪几张
3 DP：转化为少了的总和是否能有若干张卡片相加得到，dp[j+a[i]] += dp[j];
4 记录一次路径，当第一次更新的时候
5 */
6 #include <cstdio>
7 #include <iostream>
8 #include <algorithm>
9 #include <cmath>
10 #include <cstring>
11 #include <vector>
12 #include <stack>
13 using namespace std;
14
15 const int MAXN = 1e5 + 10;
16 const int INF = 0x3f3f3f3f;
17 int a[MAXN];
18 int dp[MAXN];
19 int rt[MAXN];
20
21 int main(void) //URAL 1244 Gentlemen
22 {
23 //freopen ("U.in", "r", stdin);
24
25 int tot, n;
26 while (scanf ("%d", &tot) == 1)
27 {
28 scanf ("%d", &n);
29 int sum = 0;
30 for (int i=1; i<=n; ++i) {scanf ("%d", &a[i]); sum += a[i];}
31 if (tot > sum) {puts ("0"); continue;}
32
33 memset (rt, 0, sizeof (rt));
34 memset (dp, 0, sizeof (dp));
35 sum -= tot;
36 dp[0] = 1;
37 for (int i=1; i<=n; ++i)
38 {
39 for (int j=sum-a[i]; j>=0; --j)
40 {
41 if (dp[j] && dp[j+a[i]] == 0)
42 {
43 rt[j+a[i]] = i;
44 }
45 dp[j+a[i]] += dp[j];
46 }
47 }
48
49 if (dp[sum] == 0) {puts ("0"); continue;}
50 else if (dp[sum] == 1)
51 {
52 vector<int> ans;
53 for (int i=n; i>=1; --i)
54 {
55 if (rt[sum] == i) {ans.push_back (i); sum -= a[i];}
56 }
57 for (int i=ans.size ()-1; i>=0; --i) printf ("%d%c", ans[i], (i==0) ? '
' : ' ');
58 }
59 else puts ("-1");
60 }
61
62 return 0;
63 }