1251-zlly has a steamed stuffed bun face

1251-zlly has a steamed stuffed bun face

• Time limit:1000MS
• Memory limit:65535KB

Descriptionzlly has a steamed stuffed bun face, And he especially likes to eat candy. Now he has several types of candy on his hand. Each type of candy has a taste value, and there are countless types of candy. Then, Lou's shoes also like candy very much. He has a wide variety of tastes. He asked zlly to use the existing taste values to create new taste values. Now, Lou Tong's shoes want to know the maximum taste he cannot eat? Can you help him?
For example, zlly has three types of candy on hand: 3, 6, and 10. It cannot spell out the values of 1, 2, 4, 5, and 7... 17 candy, so the result is 17.
In addition, the number of different types of candy <= 10, and the taste value of each type of candy <= 265 can ensure that the maximum result is no more than 2,000,000,000. If both can be spelled out or the maximum value does not exist, the output is 0. Inputline 1: Number of Sweets n
Line 2. n + 1: taste of various sweets;
Input files have the maximum number of flavors that cannot be spelled out by caseoutput. sample input

 
33610
 

Sample output

 
17
 

Authoradministrator sourcethe first ACM-ICPC Nanjing Invitational Tournament

 

If there are two numbers P, Q, and gcd (Q, P) = 1, the maximum value cannot be expressed as px + Qy (x> = 0, Y> = 0) the number of values is PQ-Q-P (for N> PQ-Q-P, all values can be expressed as px + Qy, while PQ-Q-P, it cannot be expressed as px + Qy ).

PS: http://www.nocow.cn/index.php/USACO/nuggets

 
# Include <cstring> # include <memory> # include <algorithm> # include <cstdio> using namespace STD; # define Max 265*265 bool DP [Max + 1]; int num [11]; int gcd (int A, int B) {if (a <B) Swap (a, B); int temp = B; if (a % B) {temp = A; A = B; B = TEMP % B;} return B;} int main () {int N, I, J, K; while (scanf ("% d", & N )! =-1) {for (I = 1; I <= N; I ++) scanf ("% d", & num [I]); k = num [1]; for (I = 2; I <= N; I ++) k = gcd (K, num [I]); If (K! = 1) {printf ("0 \ n"); continue;} memset (DP, 0, sizeof (DP); DP [0] = true; for (I = 1; I <= max; I ++) {for (j = 1; j <= N; j ++) if (I> = num [J]) DP [I] = DP [I] | DP [I-num [J];} int res = 0; for (I = max; I> = 1; I --) if (! DP [I]) {res = I; break;} printf ("% d \ n", Res);} return 0 ;}