2019牛客多校7H

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###name
Pair

###descirption
Given three integers A, B, C. Count the number of pairs <x,y> (with
1≤x≤A and 1≤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)

###input
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.
$1≤A,B,C≤10^9$

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

###sample input
3
3 4 2
4 5 2
7 8 5

###sample output
5
7
31

###toturial
可以直接dfs搜索,然后记忆化加速,写起来很复杂,但是能过

###code

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#include<bits/stdc++.h>
using namespace std;

typedef long long ll;

typedef pair<pair<int,int>,pair<int,int>> pair4;
inline pair4 mp(int a,int b,int c,int d){return make_pair(make_pair(a,b),make_pair(c,d));}

map<pair4,ll> hashxor,hashand;
ll gxor(ll a,ll b,ll hi,ll c){// ^<c
if(a==-1||b==-1)return 0;
if(hi<=1) {
ll ct=0;
for(ll i=0;i<=a;i++){
for(ll j=0;j<=b;j++){
if((i^j)<c) ct++;
}
}
return ct;
}

if(hashxor.find(mp(a,b,hi,c))!=hashxor.end()) return hashxor[mp(a,b,hi,c)];

ll a0=a>=hi-1?hi-1:a;// bg with 0
ll b0=b>=hi-1?hi-1:b;// bg with 1
ll a1=a>=hi?(a&(hi-1)):-1;
ll b1=b>=hi?(b&(hi-1)):-1;
if(c&hi) return hashxor[mp(a,b,hi,c)]=(a0+1)*(b0+1)+(a1+1)*(b1+1)+gxor(a0,b1,hi>>1,c&(hi-1))+gxor(a1,b0,hi>>1,c&(hi-1));
else return hashxor[mp(a,b,hi,c)]=gxor(a0,b0,hi>>1,c&(hi-1))+gxor(a1,b1,hi>>1,c&(hi-1));
}

ll gand(ll a,ll b,ll hi,ll c){// &>c
if(a==-1||b==-1)return 0;
if(hi<=1) {
ll ct=0;
for(ll i=0;i<=a;i++){
for(ll j=0;j<=b;j++){
if((i&j)>c) ct++;
}
}
return ct;
}

if(hashand.find(mp(a,b,hi,c))!=hashand.end()) return hashand[mp(a,b,hi,c)];

ll a0=a>=hi-1?hi-1:a;// bg with 0
ll b0=b>=hi-1?hi-1:b;// bg with 1
ll a1=a>=hi?(a&(hi-1)):-1;
ll b1=b>=hi?(b&(hi-1)):-1;
if(c&hi) return hashand[mp(a,b,hi,c)]=gand(a1,b1,hi>>1,c&(hi-1));
else return hashand[mp(a,b,hi,c)]=(a1+1)*(b1+1)+gand(a0,b1,hi>>1,c&(hi-1))+gand(a1,b0,hi>>1,c&(hi-1))+gand(a0,b0,hi>>1,c&(hi-1));
}

ll f(ll a,ll b,ll hi,ll c){// &>c ^<c
if(hi<=1){
ll ct=0;
for(ll i=0;i<=a;i++){
for(ll j=0;j<=b;j++){
if((i&j)>c||(i^j)<c) ct++;
}
}
return ct;
}
ll a0=a>=hi-1?hi-1:a;// bg with 0
ll b0=b>=hi-1?hi-1:b;// bg with 1
ll a1=a>=hi?(a&(hi-1)):-1;
ll b1=b>=hi?(b&(hi-1)):-1;
if(c&hi) return (a0+1)*(b0+1)+(a1+1)*(b1+1)+gxor(a0,b1,hi>>1,c&(hi-1))+gxor(a1,b0,hi>>1,c&(hi-1));// ^ ^ & &
else return (a1+1)*(b1+1)+gand(a0,b1,hi>>1,c&(hi-1))+gand(a1,b0,hi>>1,c&(hi-1))+f(a0,b0,hi>>1,c&(hi-1));
}


ll debug(ll a,ll b,ll c){
hashxor.clear();
hashand.clear();

ll hi=max({a,b,c});
while(hi&(hi-1)) hi&=hi-1;
return f(a,b,hi,c)+1-min(a+1,c)-min(b+1,c);
}

ll baoli(ll a,ll b,ll c){
ll ct=0;
for(ll i=1;i<=a;i++){
for(ll j=1;j<=b;j++){
if((i&j)>c||(i^j)<c) ct++;
}
}
return ct;
}

int main(){
// srand(time(NULL));
// int up=300;
// while(true){
// int i=rand()%20000+1;
// int j=rand()%20000+1;
// int k=rand()%20000+1;
// int fuck1=baoli(i,j,k);
// int fuck2=debug(i,j,k);
// cout<<i<<" "<<j<<" "<<k<<" "<<" "<<fuck1<<" "<<fuck2<<endl;
// if(fuck1!=fuck2){
// cout<<baoli(i,j,k)<<endl;
// cout<<debug(i,j,k)<<endl;
// cout<<i<<j<<k<<endl;
// getchar();
// }
//
//
// }
ll a,b,c,t;
scanf("%lld",&t);
while(t--){
hashxor.clear();
hashand.clear();

scanf("%lld%lld%lld",&a,&b,&c);
ll hi=max({a,b,c});
while(hi&(hi-1)) hi&=hi-1;
printf("%lld\n",f(a,b,hi,c)+1-min(a+1,c)-min(b+1,c));
}
}

###toturial2
考虑数位dp

###code

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#include<bits/stdc++.h>
using namespace std;
typedef long long ll;

#define REP(i,j,k) for(int i=(j);i<=(k);i++)
ll dp[33][2][2][3][3],A[33],B[33],C[33];

ll cmp(ll a,ll b){
if(a<b) return -1;
if(a==b) return 0;
return 1;
}

ll dfs(ll bit,ll la,ll lb,ll ad,ll xr){
if(bit==-1) return ad==1||xr==-1;
ll&res=dp[bit][la][lb][ad+1][xr+1];
if(res!=-1) return res;
res=0;
REP(i,0,la?A[bit]:1) REP(j,0,lb?B[bit]:1) res+=dfs(bit-1,la&&i==A[bit],lb&&j==B[bit],ad?ad:cmp(i&j,C[bit]),xr?xr:cmp(i^j,C[bit]));
return res;
}

int main(){
int t;
scanf("%d",&t);
while(t--){
memset(dp,-1,sizeof(dp));
ll a,b,c; scanf("%lld%lld%lld",&a,&b,&c);
REP(i,0,30) A[i]=bool(1<<i&a),B[i]=bool(1<<i&b),C[i]=bool(1<<i&c);
printf("%lld\n",dfs(30,1,1,0,0)+1-min(a+1,c)-min(b+1,c));
}
}

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