HDU 1085 holding bin-Laden captive! (DP)

Question link: http://acm.hdu.edu.cn/showproblem.php? PID = 1, 1085

Solution Report: There are three kinds of coins with nominal values: 1, 2, and 5. The numbers of these three coins are num_1, num_2, and num_5. Ask what is the minimum value of the money you cannot gather.

DP, open an array like DP [I] [3], then DP [I] [0] indicates that DP [I] [0] is required for making I RMB, and DP [I] [1] is required, DP [I] [2] Five blocks, and then follow the recurrence, there are three ways to obtain I, the first is the I-1 plus a piece, the second is the I-2 plus one piece of two, the fourth is the I-5 plus a five, and then each time you judge whether there is more than a number on the line, if we find that none of the three methods can be I, it means that the smallest value that cannot be I.

1 # include <cstdio> 2 3 struct node 4 {5 int A, B, C; 6} DP [8005]; 7 int main () 8 {9 int N1, N2, n5; 10 while (scanf ("% d", & N1, & N2, & N5), N1 + N2 + N5) 11 {12 DP [0]. a = 0; 13 DP [0]. B = 0; 14 DP [0]. C = 0; 15 int ans = 0; 16 for (INT I = 1; I <= N1 + 2 * N2 + 5 * N5 + 1; ++ I) // The addition of 1 is in all can be achieved in the case of direct results than the total number of 1 17 {18 if (I> = 1 & DP [I-1]. A + 1 <= N1) 19 DP [I] = DP [I-1], DP [I]. A + = 1; 20 else if (I> = 2 & DP [I-2]. B + 1 <= N2) 21 DP [I] = DP [I-2], DP [I]. B + = 1; 22 else if (I> = 5 & DP [I-5]. c + 1 <= N5) 23 DP [I] = DP [I-5], DP [I]. c + = 1; 24 else25 {26 ans = I; 27 break; 28} 29} 30 printf ("% d \ n", ANS); 31} 32 return 0; 33} 34 35

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HDU 1085 holding bin-Laden captive! (DP)