大意: 给定一个$1e9\times 1e9$的矩阵$a$, $a_{i,j}$为它正上方和正左方未出现过的最小数, 每个询问求一个矩形内的和.
可以发现$a_{i,j}=(i-1)\oplus (j-1)+1$, 暴力数位$dp$即可
#include <iostream> #include <sstream> #include <algorithm> #include <cstdio> #include <cmath> #include <set> #include <map> #include <queue> #include <string> #include <cstring> #include <bitset> #include <functional> #include <random> #define REP(i,a,n) for(int i=a;i<=n;++i) #define PER(i,a,n) for(int i=n;i>=a;--i) #define hr putchar(10) #define pb push_back #define lc (o<<1) #define rc (lc|1) #define mid ((l+r)>>1) #define ls lc,l,mid #define rs rc,mid+1,r #define x first #define y second #define io std::ios::sync_with_stdio(false) #define endl ‘\n‘ #define DB(a) ({REP(__i,1,n) cout<<a[__i]<<‘,‘;hr;}) using namespace std; typedef long long ll; typedef pair<int,int> pii; const int P = 1e9+7, INF = 0x3f3f3f3f; ll gcd(ll a,ll b) {return b?gcd(b,a%b):a;} ll qpow(ll a,ll n) {ll r=1%P;for (a%=P;n;a=a*a%P,n>>=1)if(n&1)r=r*a%P;return r;} ll inv(ll x){return x<=1?1:inv(P%x)*(P-P/x)%P;} inline int rd() {int x=0;char p=getchar();while(p<‘0‘||p>‘9‘)p=getchar();while(p>=‘0‘&&p<=‘9‘)x=x*10+p-‘0‘,p=getchar();return x;} //head //求 0<=i<=x, 0<=j<=y, 0<=i^j<=k的所有i^j的和 int f[33][2][2][2], g[33][2][2][2]; void add(int &a, ll b) {a=(a+b)%P;} //f为个数, g为和 int calc(int x, int y, int k) { if (x<0||y<0) return 0; memset(f,0,sizeof f); memset(g,0,sizeof g); f[32][1][1][1] = 1; PER(i,0,31)REP(a1,0,1)REP(a2,0,1)REP(a3,0,1) { int &r = f[i+1][a1][a2][a3]; if (!r) continue; int l1=a1&&!(x>>i&1),l2=a2&&!(y>>i&1),l3=a3&&!(k>>i&1); REP(b1,0,1)REP(b2,0,1) { if (l1&&b1) continue; if (l2&&b2) continue; if (l3&&(b1^b2)) continue; int c1=a1&&b1>=(x>>i&1),c2=a2&&b2>=(y>>i&1),c3=a3&&(b1^b2)>=(k>>i&1); add(f[i][c1][c2][c3],r); add(g[i][c1][c2][c3],g[i+1][a1][a2][a3]*2ll+(b1^b2)*r); } } int ans = 0; REP(a1,0,1)REP(a2,0,1)REP(a3,0,1) add(ans,g[0][a1][a2][a3]+f[0][a1][a2][a3]); return ans; } int main() { int t; scanf("%d", &t); while (t--) { int x1,y1,x2,y2,k; scanf("%d%d%d%d%d",&x1,&y1,&x2,&y2,&k); --x1,--y1,--x2,--y2,--x1,--y1,--k; int ans = (calc(x2,y2,k)-calc(x1,y2,k)-calc(x2,y1,k)+calc(x1,y1,k))%P; if (ans<0) ans += P; printf("%d\n", ans); } }
原文地址:https://www.cnblogs.com/uid001/p/11607126.html
时间: 2024-10-19 10:49:21