递推DP URAL 1244 Gentlemen

 

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) ? '\n' : ' '); 58  } 59 else puts ("-1"); 60  } 61 62 return 0; 63 }