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吐槽洛谷難度標簽qwq

Solution

顯然是一道神奇的DP，由於總錢數不變，我們只需要枚舉前兩個人的錢數就可知第三個人的錢數
DP的時候先枚舉只用前k個幣種，然後枚舉前兩個人的錢數，然後枚舉轉移即可

Code

#include <cmath>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <iostream>
#include <algorithm>
#define F(i,a,b) for(register int i=(a);i<=(b);i++)
using namespace std;

inline int read() {
    int x=0,f=1;char c=getchar();
    while(!isdigit(c)) {if(c=='-')f=-f;c=getchar();}
    while(isdigit(c)) x=(x<<1)+(x<<3)+c-48,c=getchar();
    return x*f;
}

const int N=1010,INF=0x3f3f3f3f;
int p[7]={0,100,50,20,10,5,1};
int sum,s1,s2,x1,x2,x3,ans;
int f[7][N][N],num[4][7];

int main() {
    x1=read(),x2=read(),x3=read();
    F(i,1,3) F(j,1,6) sum+=(num[i][j]=read())*p[j];
    F(i,1,6) s1+=num[1][i]*p[i],s2+=num[2][i]*p[i];
    if(s1-x1+x3>sum||s2-x2+x1>sum||sum-s1-s2-x3+x2>sum) 
        return puts("impossible"),0;
    memset(f,0x3f,sizeof(f));f[0][s1][s2]=0;
    F(k,1,6) {
        int tot=num[1][k]+num[2][k]+num[3][k];
        F(i,0,sum) F(j,0,sum-i) {
            if(f[k-1][i][j]==INF) continue;
            F(x,0,tot) F(y,0,tot-x) {
                int z=tot-x-y;
                int s=(abs(num[1][k]-x)+abs(num[2][k]-y)+abs(num[3][k]-z))/2;
                int d1=i-(num[1][k]-x)*p[k],d2=j-(num[2][k]-y)*p[k];
                if(d1+d2>sum) continue; if(d1<0||d2<0) continue;
                f[k][d1][d2]=min(f[k][d1][d2],f[k-1][i][j]+s);
            }
        }
    } 
    int ans=f[6][s1-x1+x3][s2-x2+x1];
    if(ans==INF) return puts("impossible"),0;
    return printf("%d",ans),0;
}
 

[luogu4026 SHOI2008]循環的債務 (DP)