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);}