洛谷 P5020 【货币系统】

"简" "单" 的完全背包O_O

来一个开心的AC代码

#include<cstdio>
#include<cstring>
#define __________ 100005
#define ___________ "%d"
#define ____________ scanf
#define _____________ memset
#define ______________ int
#define _______________ for
#define ________________ while
#define _________________ inline 
#define __________________ return
#define ___________________ main
#define ____________________ for
#define _____________________ if
#define ______________________ sizeof
#define _______________________ using 
#define ________________________ namespace
#define _________________________ std 
#define __________________________ printf
#define ______________________________________________________ "%d\n"
_______________________ ________________________ _________________________;

______________ _____,________[__________],___[__________],____=-1,_;

_________________ ______________ __(______________ ___________________________,______________ ____________________________){
    __________________ ___________________________>____________________________?___________________________:____________________________;
}

______________ ___________________(){
    ____________(___________,&_);
    ________________(_--){
        ____=-1;
        ______________ ______=0;
        _____________(___,0,______________________(___));
        ____________(___________,&_____);
        ___[0]=1;
        ____________________(______________ _______=1;_______<=_____;_______++){
            ____________(___________,&________[_______]);
            ____=__(____,________[_______]);
        }
        ____________________(______________ _______=1;_______<=_____;_______++){
            ____________________(______________ _________=________[_______];_________<=____;_________++){
                _____________________(___[_________-________[_______]])___[_________]++;
            }
        }
        ____________________(______________ _______=1;_______<=_____;_______++){
            _____________________(___[________[_______]]>1){
                ______++;
            }
        }
        __________________________(______________________________________________________,_____-______);
    }
}

好吧他本来长这样

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

int n,a[100005],dp[100005],maxx=-1;

inline int max(int x,int y){
    return x>y?x:y;
}

int main(){
    int t;
    scanf("%d",&t);
    while(t--){
        maxx=-1;
        int ans=0;
        memset(dp,0,sizeof(dp));
        scanf("%d",&n);
        dp[0]=1;
        for(int i=1;i<=n;i++){
            scanf("%d",&a[i]);
            maxx=max(maxx,a[i]);
        }
        for(int i=1;i<=n;i++){
            for(int j=a[i];j<=maxx;j++){
                if(dp[j-a[i]])dp[j]++;
            }
        }
        for(int i=1;i<=n;i++){
            if(dp[a[i]]>1){
                ans++;
            }
        }
        printf("%d\n",n-ans);
    }
}

• 根据题意，如果某一种面额可以被其他钱币拼凑而成，那么它就可以舍去。

• 先dp出无穷多钱币可以拼凑出的钱币种类(小于最大面额即可)。如果一种钱数j可以被拼凑出，我们令dp[j]=1.

• 然后枚举提供的每种面额，如果dp值为1，那么ans(可以被舍去的面额的数量)++。

• 最后输出n-ans即可。

//然而考试的时候我完全背包打错了O_O惊吓，结果居然强势AC O_O

转载于:https://www.cnblogs.com/Y15BeTa/p/luogu5020.html