【BZOJ3107】二进制a+b，DP

Time:2016.08.24
Author:xiaoyimi
转载注明出处谢谢

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传送门
思路：
今天的模拟题之一
比较奥妙重重的DP思路
f[i][j][k][l][0/1]表示DP到了第i位
此时X有j个1，Y有k个1，Z有l个1
i+1位是0还是1
f[i][j][k][l][0]−>⎧⎩⎨⎪⎪⎪⎪⎪⎪f[i+1][j+1][k+1][l+1][1]f[i+1][j+1][k][l+1][0]f[i+1][j][k+1][l+1][0]f[i+1][j][k][l][0]

f[i][j][k][l][1]−>⎧⎩⎨⎪⎪⎪⎪⎪⎪f[i+1][j+1][k+1][l+1][1]f[i+1][j][k+1][l][1]f[i+1][j+1][k][l][1]f[i+1][j][k][l][0]
答案就是f[n][a的1个数][b的1个数][c的1个数][0]
转移复杂度为O(log42n)常数略大
注意：要开一些奥妙重重的longlong
联动Shallwe的O(log2n)构造法
折越
代码：

#include<cstdio>#include<iostream>#include<cstring>#include<algorithm>#include<cmath>#define LL long longusing namespace std;int T,a,b,c;LL f[33][33][33][33][2];int cal(int x){    int sum=0;    for (;x;x>>=1)        sum+=x&1;    return sum;}void work(){    scanf("%d%d%d",&a,&b,&c);    memset(f,127,sizeof(f));    int n=max((int)log2(a)+1,(int)log2(b)+1);    n=max(n,(int)log2(c)+1);    int la=cal(a),lb=cal(b),lc=cal(c);    f[0][0][0][0][0]=0;    for (int i=0;i<n;++i)        for (int j=0;j<=la;++j)            for (int k=0;k<=lb;++k)                for (int l=0;l<=lc;++l)                {                    LL tmp=f[i][j][k][l][0];                    f[i+1][j+1][k+1][l+1][1]=min(f[i+1][j+1][k+1][l+1][1],tmp+(1<<i+1));                    f[i+1][j+1][k][l+1][0]=min(f[i+1][j+1][k][l+1][0],tmp+(1<<i));                    f[i+1][j][k+1][l+1][0]=min(f[i+1][j][k+1][l+1][0],tmp+(1<<i));                    f[i+1][j][k][l][0]=min(f[i+1][j][k][l][0],tmp);                    tmp=f[i][j][k][l][1];                    f[i+1][j+1][k+1][l+1][1]=min(f[i+1][j+1][k+1][l+1][1],tmp+(1<<i+1));                    f[i+1][j][k+1][l][1]=min(f[i+1][j][k+1][l][1],tmp+(1<<i));                    f[i+1][j+1][k][l][1]=min(f[i+1][j+1][k][l][1],tmp+(1<<i));                    f[i+1][j][k][l][0]=min(f[i+1][j][k][l][0],tmp);                }    printf("%lld\n",f[n][la][lb][lc][0]>=(1LL<<31)-1?-1:f[n][la][lb][lc][0]);}main(){work();}