"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。
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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.
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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)++。
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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