牛客多校第六场-H-Pair

链接：https://ac.nowcoder.com/acm/contest/887/H
来源：牛客网

题目描述

Given three integers A, B, C. Count the number of pairs <x ,y> (with 1≤x≤Aand1≤y≤B)
such that at least one of the following is true:
- (x and y) > C
- (x xor y) < C
 
("and", "xor" are bit operators)

输入描述:

The first line of the input gives the number of test cases,T(T≤100). T test cases follow.

For each test case, the only line contains three integers A, B and C.
1≤A,B,C≤10^9

输出描述:

For each test case, the only line contains an integer that is the number of pairs satisfying the condition given in the problem statement.

示例1

输入

3
3 4 2
4 5 2
7 8 5

输出

5
7
31

数位dp求 x&y<=c && x^y>=c的个数然后用所有方案剪掉
具体见代码

#include <bits/stdc++.h>
#define ll long long
using namespace std;
int T,A,B,C;
int a[],b[],c[];
ll f[][][][][];
void cal(int x,int v[])
{
    for (int i=;i<=;i++) v[i]=(x>>i)&;
}
ll dfs(int pos,bool lima,bool limb,bool limand,bool limxor)
//          位       x<a     y<b         x&y<c       x^y>c
{
    if (pos==-) return ;
    if (f[pos][lima][limb][limand][limxor]!=-) return f[pos][lima][limb][limand][limxor];

    int aa=lima?a[pos]:;   //lima=1说明之前一直相等当前位取要<=a，否则说明之前<a了当前位可以乱取
    int bb=limb?b[pos]:;
    int c1=limand?c[pos]:; //同理如果之前x&y一直等于c那么当前位要x&y<=c，否则就可以乱取
    int c2=limxor?c[pos]:;
    ll &ret=f[pos][lima][limb][limand][limxor];
    ret=;

    for (int i=;i<=aa;i++)
    for (int j=;j<=bb;j++)
    {
        if ((i&j)>c1) continue;
        if ((i^j)<c2) continue;
        ret+=dfs(pos-,lima&&i==aa,limb&&j==bb,limand&&(i&j)==c1,limxor&&(i^j)==c2);
    }
    return ret;
}
int main()
{
    scanf("%d",&T);
    while(T--)
    {
        scanf("%d%d%d",&A,&B,&C);
        memset(a,,sizeof(a));
        memset(b,,sizeof(b));
        memset(c,,sizeof(c));
        memset(f,-,sizeof(f));
        cal(A,a); cal(B,b); cal(C,c);
        ll ans=dfs(,,,,);
        ans-=max(,A-C+);
        ans-=max(,B-C+);  //减掉x,y为0的情况
        printf("%lld\n",(ll)A*B-ans);

    }
    return ;
}