Gym 100963B

Gym 100963B
啊，郁闷，就tm调小了一点范围就A了，就写dp和贪心比较一下，范围到最大值的二倍-1就好了
假设最大值的2倍以内能满足最优条件，当金额范围超过最大值2倍的时候；
至于为什么，还不清楚，再想想

#include<iostream>
#include<cstdio>
#include<queue>
#include<algorithm>
#include<cmath>
#include<ctime>
#include<set>
#include<map>
#include<stack>
#include<cstring>
#define inf 2147483647
#define ls rt<<1
#define rs rt<<1|1
#define lson ls,nl,mid,l,r
#define rson rs,mid+1,nr,l,r
#define N 100010
#define For(i,a,b) for(register int i=a;i<=b;i++)
#define p(a) putchar(a)
#define g() getchar()

using namespace std;
int n;
int f[10010];
int a[100];
bool flag;
int cnt;
int greedy,dp;

void in(int &x){
    int y=1;
    char c=g();x=0;
    while(c<'0'||c>'9'){
        if(c=='-')y=-1;
        c=g();
    }
    while(c<='9'&&c>='0'){
        x=(x<<1)+(x<<3)+c-'0';c=g();
    }
    x*=y;
}
void o(int x){
    if(x<0){
        p('-');
        x=-x;
    }
    if(x>9)o(x/10);
    p(x%10+'0');
}

void clear(){
    For(i,1,10000){
        f[i]=inf;
    }
    greedy=0;
}

int get1(int x){
    int r=0;
    for(int i=n;i;i--){
        r+=x/a[i];
        x%=a[i];
    }
    return r;
}

int get2(int x){
    //f[0]=1;
    For(i,1,n)
        For(j,a[i],x)
            f[j]=min(f[j],f[j-a[i]]+1);
    return f[x];
}

int main(){
    while(cin>>n&&n!=0){
        For(i,1,n)
            in(a[i]);
        if(a[1]!=1){
            cout<<"Case #"<<++cnt<<": Cannot pay some amount"<<endl;
            continue;
        }
        clear();
        For(i,1,a[n]*2-1){
            greedy=get1(i);
            dp=get2(i);
            if(greedy!=dp){
                cout<<"Case #"<<++cnt<<": Cannot use greedy algorithm"<<endl;
                //cout<<i<<" "<<greedy<<" "<<dp<<endl;
                break;
            }
        }
        if(greedy!=dp) continue;
        cout<<"Case #"<<++cnt<<": OK"<<endl;
    }
    return 0;
}