HDU 1085 Holding Bin-laden captive! The basic use of the mother function, or is not difficult

Holding Bin-laden captive!Time limit:2000/1000 MS (java/others) Memory limit:65536/32768 K (java/others)
Total submission (s): 16063 Accepted Submission (s): 7206


Problem descriptionwe all know this bin-laden is a notorious terrorist, and he have disappeared for a long time. But recently, it's reported that he hides in hang Zhou of china!
"Oh, god! How terrible! ”



Don ' t is so afraid, guys. Although he hides in a cave of the Hang Zhou, the He dares not the go out. Laden is so bored recent years that he fling himself to some math problems, and he said that if anyone can solve his pro Blem, he'll give himself up!
ha-ha! Obviously, Laden is too proud of he intelligence! But, what's his problem?
"Given some Chinese Coins (coins) (three kinds--1, 2, 5), and their number is num_1, num_2 and num_5 respectively, please OU Tput the minimum value that's cannot pay with given coins. "
You, super Acmer, should solve the problem easily, and don ' t forget to take $25000000 from bush!

Inputinput contains multiple test cases. Each test case contains 3 positive integers num_1, num_2 and Num_5 (0<=num_i<=1000). A test case containing 0 0 0 terminates the-input and this-test case are not-to-be processed.

Outputoutput the minimum positive value that one cannot pay with given coins, one line for one case.

Sample Input

1 1 30) 0 0

Sample Output

4

If you have just learned the function of the mother, then this problem may be a hurdle for you, do not be afraid, it is best not to look at other people's Code and problems, if this problem oneself figured out, I want to understand the mother function, will go to the next level, refueling bar code:

#include <stdio.h> #define MAX 10100int Main () {int C1[max], C2[max], num[5], a[]={0,1,2,5}; while (~SCANF ("%d%d%d", &NUM[1],&NUM[2],&NUM[3]) && (num[1]| | num[2]| | NUM[3]) {int m = num[1]*a[1]+num[2]*a[2]+num[3]*a[3]; for (int i = 0; I <= m; ++i) {c1[i] = 1; C2[i] = 0;} C1[m+1] = 0; int  len = num[1]*a[1]; for (int i = 2; I <= 3; ++i) {len + = Num[i]*a[i]; for (int j = 0; J <= Len-n Um[i]*a[i]; J + +) {for (int k = 0; k+j <= len; k+=a[i]) {C2[j+k] + = C1[j];}} for (int j = 0; J <= Len; ++j) {c1[j] = C2[j]; C2[j] = 0;}} for (int i = 0; i < MAX; ++i) {if (c1[i] = = 0) {printf ("%d\n", I); break;}}} return 0;}

with June

HDU 1085 Holding Bin-laden captive! The basic use of the mother function, or is not difficult