POJ3040

#include<cstdio>
#include<algorithm>
#include<cstring>
using namespace std;

const int maxn = 23;
int N, C;
struct node{
    int price;
    int amount;
    bool operator < (const node& n) const{
        return price < n.price;
    }
}coins[maxn];
int used[maxn];

void solve(){
    sort(coins, coins+N);
    int ans = 0;
    //第一次貪心，從大于C的開始選擇
    for(int i = N-1; i >= 0; i--)
        if(coins[i].price >= C){
            ans += coins[i].amount;
            coins[i].amount = 0;
        }
    while(true){
        bool flag = false;
        int money = C;
        memset(used, 0, sizeof(used));
        //第二次貪心首先從小于C的選擇，但是要從大開始選
        for(int i = N-1; i >= 0; i--){
            if(coins[i].amount != 0){
                int k = money / coins[i].price;
                int m = min(k, coins[i].amount);
                money -= m*coins[i].price;
                used[i] = m;
            }
            if(money == 0){
                flag = true;
                break;
            } 
        }
        if(money > 0){
            //接着從小的開始選擇
            for(int i = 0; i < N; i++){
                if(coins[i].amount > used[i]){
                    while(used[i] < coins[i].amount){
                        money -= coins[i].price;
                        used[i]++;
                        if(money <= 0){
                            flag = true; 
                            break;
                        }
                    }
                }
                if(flag){
                    break;
                }
            }
        }
        if(!flag){//如果flag為false表示剩下的錢不足以湊夠C，可以結束了
            break;
        }
        int m = 0x3f3f3f3f;
        for(int i = 0; i < N; i++){
            if(used[i])
                m = min(m, coins[i].amount;
        } 
        ans += m;
        for(int i = 0; i < N; i++){
            if(used[i])
                coins[i].amount -= m*used[i]; 
        }
    }
    printf("%d\n", ans);
}

int main(){
    scanf("%d%d", &N, &C);  
    for(int i = 0; i < N; i++)
        scanf("%d%d", &coins[i].price, &coins[i].amount);
    solve();
    return 0;
}