Holding Bin-Laden Captive! HDU-1085 母函数

HDU-1085

Don’t be so afraid, guys. Although he hides in a cave of Hang Zhou, he dares not to go out. Laden is so bored recent years that he fling himself into some math problems, and he said that if anyone can solve his problem, he will give himself up!

Ha-ha! Obviously, Laden is too proud of his intelligence! But, what is his problem?

“Given some Chinese Coins (硬币) (three kinds-- 1, 2, 5), and their number is num_1, num_2 and num_5 respectively, please output the minimum value that you cannot pay with given coins.”

You, super ACMer, should solve the problem easily, and don’t forget to take $25000000 from Bush!

input

1 1 3

0 0 0

output

4

code

//Siberian Squirrel
//#include<bits/stdc++.h>
#include<unordered_map>
#include<algorithm>
#include<iostream>
#include<cstring>
#include<cmath>

#define ACM_LOCAL

using namespace std;
typedef long long ll;

const double PI = acos(-1);
const double eps = 1e-7;
const int MOD = 3221225473;
const int N = 5e4 + 10;
const int UP = 1e4;

int a[4], b[4] = {0, 1, 2, 5};
bool f[N];

inline ll solve(ll res = 0) {
    memset(f, false, sizeof f);
    f[0] = true;
    for(int i = 1; i <= 3; ++ i) {
        for(int j = 10000; j >= 0; -- j) {
            for(int k = 0; k <= a[i]; ++ k) {
                if(f[j]) {
                    f[j + k * b[i]] = true;
                }
            }
        }
    }
    for(int i = 1; i <= 10000; ++ i) if(!f[i]) return i;
}

int main() {
#ifdef ACM_LOCAL
    freopen("input", "r", stdin);
    freopen("output", "w", stdout);
#endif
    int o = 1, n, m;
//    scanf("%d", &o);
    while(o --) {
        while(~scanf("%d%d%d", &a[1], &a[2], &a[3]) && a[1] + a[2] + a[3])
            printf("%lld\n", solve());
    }
    return 0;
}