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思路：从起始点开始，每隔（2^n）*（2^n）（n为任意正整数）矩阵之间，不同矩阵中数的和之差为n*n*n，而连续个x个（2^n）*m矩阵之和为(1+x*2^n)*x*2^n/2，因此我们可以快速统计连续2^n区间和，对k进行切割即可。

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

#define MOD 1000000007
#define LL long long

LL qpow(LL x,LL y){
    LL e=1;
    while(y){
        if(y%2)e=(e*x)%MOD;
        x=x*x%MOD;
        y/=2;
    }
    return e;
}

LL low(LL x){

    for(LL i=1;i<=x;i<<=1){
        if(i<<1>x)
            return i;
    }
}

LL getS(LL x,LL y,LL n){
    return (y-x+1)*(x+y)%MOD*qpow(2,MOD-2)%MOD*n%MOD;
}


LL dfs(LL a,LL b,LL k,LL t){

    if(a<=0||b<=0||k<=0) return 0;

    LL l=min(a,b);
    LL r=max(a,b);

    LL x=low(l);

    LL limit=min(r/x,k/x);
    LL ans=0;

    if(l==r&&l==x){
        ans=(ans+getS(t+1,t+min(x,k),x))%MOD;
    }
    else{
        if(limit%2==0){
            LL rema=min(r-limit*x,x);
            ans=(ans+dfs(x,rema,k-limit*x,t+limit*x))%MOD;
            ans=(ans+dfs(l-x,rema,k-limit*x-x,t+limit*x+x))%MOD;
            rema=min(r-limit*x-x,x);
            ans=(ans+dfs(x,rema,k-limit*x-x,t+limit*x+x))%MOD;
            ans=(ans+dfs(l-x,rema,k-limit*x,t+limit*x))%MOD;
        }
        else{
            limit--;
            ans=(ans+dfs(x,x,k-limit*x,t+limit*x))%MOD;
            ans=(ans+dfs(l-x,x,k-limit*x-x,t+limit*x+x))%MOD;
            LL rema=min(r-limit*x-x,x);
            ans=(ans+dfs(x,rema,k-limit*x-x,t+limit*x+x))%MOD;
            ans=(ans+dfs(l-x,rema,k-limit*x,t+limit*x))%MOD;
        }
        ans=(ans+getS(t+1,t+limit*x,l))%MOD;
    }
    return ans;
}

int main() {

    int N,x1,x2,y1,y2,k;
    scanf("%d", &N);
    for(int i=1;i<=N;i++){
        scanf("%d%d%d%d%d",&x1,&y1,&x2,&y2,&k);
        LL ans=((dfs(x2,y2,k,0)-dfs(x1-1,y2,k,0)-dfs(x2,y1-1,k,0)+dfs(x1-1,y1-1,k,0))%MOD+MOD)%MOD;
        printf("%lld\n",ans);
    }
    return 0;
}