P4574 [CQOI2013]二进制A+B

P4574 [CQOI2013]二进制A+B

数位dp

设f[ i ][ a ][ b ][ c ][ 0/1 ]表示在第 n-i 位，A、B、C各贡献了a、b、c个1，是否需要进位的情况下，c的最小值

然后对于8种情况枚举

#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
typedef long long ll;
template <typename T> inline T min(T &a,T &b) {return a<b ?a:b;}
template <typename T> inline int find(T a) {int res=0; for(;a;a>>=1)++res; return res;} //计算位数
template <typename T> inline int cont(T x) {int res=0; for(;x;x&=(x-1))++res; return res;} //计算二进制下1的个数
ll f[33][33][33][33][2],ans,inf;
int main(){
    memset(f,63,sizeof(f)); inf=ans=f[0][0][0][0][0];
    int A,B,C,mxd;
    scanf("%d%d%d",&A,&B,&C);
    mxd=max(max(find(A),find(B)),find(C));
    A=cont(A),B=cont(B),C=cont(C);
    f[0][0][0][0][0]=0;
    for(int i=0;i<=mxd;++i)
        for(int a=0;a<=min(i+1,A);++a)
            for(int b=0;b<=min(i+1,B);++b)
                for(int c=0;c<=min(i+1,C);++c){ //8种枚举
                    f[i+1][a][b][c][0]=min(f[i+1][a][b][c][0],f[i][a][b][c][0]<<1);
                    f[i+1][a+1][b][c+1][0]=min(f[i+1][a+1][b][c+1][0],f[i][a][b][c][0]<<1|1);
                    f[i+1][a][b+1][c+1][0]=min(f[i+1][a][b+1][c+1][0],f[i][a][b][c][0]<<1|1);
                    f[i+1][a+1][b+1][c][0]=min(f[i+1][a+1][b+1][c][0],f[i][a][b][c][1]<<1);
                    f[i+1][a][b][c+1][1]=min(f[i+1][a][b][c+1][1],f[i][a][b][c][0]<<1|1);
                    f[i+1][a][b+1][c][1]=min(f[i+1][a][b+1][c][1],f[i][a][b][c][1]<<1);
                    f[i+1][a+1][b][c][1]=min(f[i+1][a+1][b][c][1],f[i][a][b][c][1]<<1);
                    f[i+1][a+1][b+1][c+1][1]=min(f[i+1][a+1][b+1][c+1][1],f[i][a][b][c][1]<<1|1);
                }
    for(int i=0;i<=mxd;++i) ans=min(ans,f[i][A][B][C][0]);
    if(ans==inf) printf("-1");
    else printf("%lld",ans);
    return 0;
}