[The third bullet of lamps] main function-hdu1085-AC under Taishan

Just after coming down from the Thai mountains, I felt that as long as the walking was stable and the rest was not stopped, it was a weak blow to dongtianmen or 18 disks .. Say something about half a ninety-nine in the industry ...... That's how sunrise is ...... 36 h not sleeping ~~~

It is slightly different from the template. Bin Laden scored 1, 2, and 5 points, asking him for the minimum amount of money he could not make. From this question, I finally figured out that the two repeated loops are used to evaluate the power, C1 is the first expression, C2 is the second expression, and the value of the array represents its subscript, c2 [I + J] = c1 [I] is to add the power value. There is no problem with the algorithm. The input format is stuck with three-side wa, so it should not be ......

# Include <iostream> # include <cstring> using namespace STD; const int INF = 10000; int c1 [INF + 1], C2 [INF + 1], c3 [INF + 1]; bool canchange; int money [3] = {1, 2, 5}; // number of times the original subscript is the power of the dog's day !!!! I cut the grass !! Int main () {int A, B, C; while (CIN> A> B> C & (! = 0 | B! = 0 | C! = 0) & (A + B + C> 0) // low-level error, Middle | written as &, broken eggs ...... {Memset (C1, 0, sizeof (C1); memset (C2, 0, sizeof (C2); memset (C3, 0, sizeof (C3 )); canchange = false; For (int t = 0; t <= A; t ++) {c1 [T] = 1 ;}for (INT I = 0; I <= A; I ++) {for (Int J = 0; j <= B * 2; j + = 2) {c2 [I + J] + = c1 [I] ;}} for (INT I = 0; I <= a + (2 * B); I ++) {for (Int J = 0; j <= C * 5; j + = 5) {C3 [I + J] + = c2 [I] ;}} for (INT I = 0; I <= a + 2 * B + 5 * C; I ++) {If (C3 [I] = 0) {cout <I <Endl; canchange = true; break ;}} if (canchange = false) {cout <A * 1 + B * 2 + C * 5 + 1 <Endl; // if there is no way to do this, consider} return 0 ;}

Finally, I would like to express my deep nostalgia to Comrade bin Laden .. $25000000 Let Bush digest it ......