HDU 5803 Zhu’s Math Problem

Posted 2020-11-03 LJZ_C

标签（空格分隔）： 数位DP

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题目链接

将ABCD拆成二进制位,则对于一个不等式,差大于2时一定成立,小于-2时一定不成立

#include<cstdio>
#include<cstring>
#include<iostream>
using namespace std;
typedef long long ll;
ll read()
{
    ll x=0,f=1;char ch=getchar();
    while(ch<‘0‘||ch>‘9‘){if(ch==‘-‘)f=-1;ch=getchar();}
    while(ch>=‘0‘&&ch<=‘9‘){x=x*10+ch-‘0‘;ch=getchar();}
    return x*f;
}
#define rep(a,b,c) for(int a=b;a<=c;a++)
#define calc(x,y) ((bool(s&y))|((x>>step)&1))
const int mod=1e9+7;
inline void MOD(int &x,int y){x+=y;if(x>=mod)x-=mod;}
ll A,B,C,D;int T,f[64][4][4][16];
int dp(int step,int c0,int c1,int s)
{
    if(step<0)return c0>0&&c1>=0;
    if(~f[step][c0+1][c1+1][s])return f[step][c0+1][c1+1][s];
    int x0=calc(A,1),x1=calc(B,2),x2=calc(C,4),x3=calc(D,8),res=0;
    rep(i,0,x0)rep(j,0,x1)rep(h,0,x2)rep(k,0,x3)
    {
        int t1=min(2,c0*2+i+h-j-k),t2=min(2,c1*2+i+k-j-h);
        if(t1<-1||t2<-1)continue;
        int t3=s|((i!=x0)<<0)|((j!=x1)<<1)|((h!=x2)<<2)|((k!=x3)<<3);
        MOD(res,dp(step-1,t1,t2,t3));
    }
    return f[step][c0+1][c1+1][s]=res;
}
int main()
{
    T=read();
    rep(kase,1,T)
    {
        A=read(),B=read(),C=read(),D=read();
        memset(f,-1,sizeof(f));
        printf("%d\n",dp(60,0,0,0));
    }
    return 0;
}