POJ 3040 贪心

贪心好题
……….
思路：
从大到小凑C 如果不够 再从小到大补满（超过）C

//By SiriusRen
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
int n,c,ans,flag,vis[21];
struct Money{int amount,value;}money[100];
bool cmp(Money a,Money b){return a.value<b.value;}
int main(){
    scanf("%d%d",&n,&c);
    for(int i=1;i<=n;i++)
        scanf("%d%d",&money[i].value,&money[i].amount);
    sort(money+1,money+1+n,cmp);
    while(1){
        memset(vis,0,sizeof(vis));
        int sum=c;
        for(int i=n;i;i--)
            if(sum>0&&money[i].amount){
                int temp=min(money[i].amount,sum/money[i].value);
                if(temp>0)sum-=money[i].value*temp,vis[i]+=temp;
            }
        for(int i=1;i<=n;i++)
            if(sum>0&&money[i].amount&&money[i].value>=sum){
                sum-=money[i].value;vis[i]++;break;
            }
        if(sum>0)break;
        for(int i=1;i<=n;i++)
            money[i].amount-=vis[i];
        ans++;
    }
    printf("%d\n",ans);
}

卡时过得…
然后我发现 诶呦 一次可以同时消很多 然后就0msAC了,,,,

//By SiriusRen
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
int n,c,ans,flag,vis[21];
struct Money{int amount,value;}money[100];
bool cmp(Money a,Money b){return a.value<b.value;}
int main(){
    scanf("%d%d",&n,&c);
    for(int i=1;i<=n;i++)
        scanf("%d%d",&money[i].value,&money[i].amount);
    sort(money+1,money+1+n,cmp);
    while(1){
        memset(vis,0,sizeof(vis));
        int sum=c,temp=0x3ffffff;
        for(int i=n;i;i--)
            if(sum>0&&money[i].amount){
                int temp=min(money[i].amount,sum/money[i].value);
                if(temp>0)sum-=money[i].value*temp,vis[i]+=temp;
            }
        for(int i=1;i<=n;i++)
            if(sum>0&&money[i].amount&&money[i].value>=sum){
                sum-=money[i].value;vis[i]++;break;
            }
        if(sum>0)break;
        for(int i=1;i<=n;i++){
            if(!vis[i])continue;
            temp=min(temp,money[i].amount/vis[i]);
        }
        for(int i=1;i<=n;i++){
            money[i].amount-=vis[i]*temp;
        }
        ans+=temp;
    }
    printf("%d\n",ans);
}