Sequence

Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 262144/262144 K (Java/Others)
Total Submission(s): 1731 Accepted Submission(s): 656

Problem Description

Let us define a sequence as below

F 1=A

F 2=B

F n = C \cdot F n - 2 +

Your job is simple, for each task, you should output Fn module 109+7.

Input

The first line has only one integer T, indicates the number of tasks.

Then, for the next T lines, each line consists of 6 integers, A , B, C, D, P, n.

1≤T≤200≤A,B,C,D≤1091≤P,n≤109

Sample Input

3 3 2 1 3 5

3 2 2 2 1 4

Sample Output

Source

2018 Multi-University Training Contest 7

cin>>n;
ll ans = 0;
for(ll l = 1,r;l <= n;l = r + 1){
     r = n / (n / l);
     ans += (r - l + 1) * (n / l);
}
cout<<ans<<endl;

除法分块模板

[p/n]是整除，一段内的值是相同的，他的整除值有sqrt（p）种。

因此可以将变量分块每块看作常量，对每一块使用矩阵快速幂。

#include<iostream>
#include<stdio.h>
#include<stdlib.h>
#include<string.h>
#include<math.h>
#include<algorithm>
#define MAX 10
#define INF 0x3f3f3f3f
#define MOD 1000000007
using namespace std;
typedef long long ll;

ll p,q;
struct mat{
    ll a[MAX][MAX];
};

mat operator *(mat x,mat y)
{
    mat ans;
    memset(ans.a,0,sizeof(ans.a));
    for(int i=1;i<=3;i++){
        for(int j=1;j<=3;j++){
            for(int k=1;k<=3;k++){
                ans.a[i][j]+=x.a[i][k]*y.a[k][j]%MOD;
                ans.a[i][j]%=MOD;
            }
        }
    }
    return ans;
}
mat qMod(ll x,mat a,ll n)
{
    mat t;
    t.a[1][1]=q;t.a[1][2]=p;t.a[1][3]=x;
    t.a[2][1]=1;t.a[2][2]=0;t.a[2][3]=0;
    t.a[3][1]=0;t.a[3][2]=0;t.a[3][3]=1;
    while(n){
        if(n&1) a=t*a;
        n>>=1;
        t=t*t;
    }
    return a;
}
int main()
{
    int t,i;
    ll a1,a2,x,n;
    scanf("%d",&t);
    while(t--){
        scanf("%lld%lld%lld%lld%lld%lld",&a1,&a2,&p,&q,&x,&n);
        if(n==1) printf("%lld\n",a1);
        else if(n==2) printf("%lld\n",a2);
        else{
            mat a;
            a.a[1][1]=a2;a.a[1][2]=0;a.a[1][3]=0;
            a.a[2][1]=a1;a.a[2][2]=0;a.a[2][3]=0;
            a.a[3][1]=1;a.a[3][2]=0;a.a[3][3]=0;
            if(x>=n){
                for(i=3;i<=n;i=x/(x/i)+1){
                    a=qMod(x/i,a,min(n,x/(x/i))-i+1);
                }
            }
            else{
                for(i=3;i<=x;i=x/(x/i)+1){
                    a=qMod(x/i,a,x/(x/i)-i+1);
                }
                a=qMod(0,a,n-max(x,2ll));
            }
            printf("%lld\n",a.a[1][1]);
        }
    }
    return 0;
}

来源：oschina

链接：https://my.oschina.net/u/4409418/blog/3865652

标签

Sequence

java

HDU-6395 多校7 Sequence（除法分块+矩阵快速幂）

Sequence