[数位DP] Codeforces 809C Round #415 (Div. 1) C. Find a car

这个东西就是(x−1) xor (y−1)+1
然后转化为 ∑i∑j[i xor j≤k]和∑i∑j[i xor j≤k]∗(i xor j)
这个东西 数位dp下就好了
代码常数略大 但是还是很好看的

#include<cstdio>#include<cstdlib>#include<algorithm>#include<cstring>#define cl(x) memset(x,0,sizeof(x))using namespace std;typedef long long ll;#define read(x) scanf("%d",&(x))const int P=1e9+7;ll sum[35][2][2][2],cnt[35][2][2][2];inline ll Calc(int _x,int _y,int _K){  if (_x==-1 || _y==-1) return 0;  int x[35]={0},y[35]={0},K[35]={0},_p=0;  _p=-1; while (_x) x[++_p]=_x&1,_x>>=1;  _p=-1; while (_y) y[++_p]=_y&1,_y>>=1;  _p=-1; while (_K) K[++_p]=_K&1,_K>>=1;  cl(sum); cl(cnt);  cnt[31][0][0][0]=1;  for (int i=31;i;i--)    for (int t1=0;t1<2;t1++)      for (int t2=0;t2<2;t2++)    for (int t3=0;t3<2;t3++){      for (int a=0;a<=(t1?1:x[i-1]);a++)        for (int b=0;b<=(t2?1:y[i-1]);b++){          if ((a^b)>(t3?1:K[i-1])) continue;          int nt1=t1|(a<x[i-1]),nt2=t2|(b<y[i-1]),nt3=t3|((a^b)<K[i-1]);          (cnt[i-1][nt1][nt2][nt3]+=cnt[i][t1][t2][t3])%=P;          (sum[i-1][nt1][nt2][nt3]+=sum[i][t1][t2][t3]+cnt[i][t1][t2][t3]*((a^b)<<(i-1))%P)%=P;        }    }  ll Sum=0,Cnt=0;  for (int t1=0;t1<2;t1++)    for (int t2=0;t2<2;t2++)      for (int t3=0;t3<2;t3++)    (Sum+=sum[0][t1][t2][t3])%=P,(Cnt+=cnt[0][t1][t2][t3])%=P;  return (Sum+Cnt)%P;}int main(){  int Q,x1,x2,y1,y2,K;  freopen("t.in","r",stdin);  freopen("t.out","w",stdout);  read(Q);  while (Q--){    read(x1); read(y1); read(x2); read(y2); read(K); x1--; y1--; x2--; y2--; K--;    printf("%I64d\n",(Calc(x2,y2,K)+P-Calc(x2,y1-1,K)+P-Calc(x1-1,y2,K)+Calc(x1-1,y1-1,K))%P);  }  return 0;}