Codeforces 1166

1166 D

The meaning of problems

We say that a sequence \ (X \) (length \ (n-\) ) is \ (m-cute \) if and only if \ (x_i = x_ {i - 1} + x_ {i - 2} + \ x_1 + + r_i DOTS (. 1 \ Le r_i \ Le m) \) . Now given \ (x \) of the first term and last term and \ (m \) , and asked whether there is a legitimate \ (x \) . There is also a need to construct. \ ((x_1, x_n, m \ le 10 ^ {14}) \)

Example

input
2
5 26 2
3 9 1
output
4 5 6 13 26
-1

solution

First assumes that all \ (= r_i. 1 \) , was calculated \ (n-\) , then add up one by one from left to right

Code

#include<bits/stdc++.h>
using namespace std;
const int maxn=53;
long long a[maxn],c[maxn],s,t,m;
int n;
void calcn(){
    n=2;
    long long sum=s;
    a[1]=s;
    while(1){
        a[n]=sum+1;
        if(a[n]>t)break;
        sum+=a[n];
        n++;
    }
    n--;
}
int main(){
    int T;
    scanf("%d",&T);
    while(T--){
        scanf("%lld%lld%lld",&s,&t,&m);
        calcn();
        for(int i=1;i<=n;i++)c[i]=0;
        long long x=t-a[n],y,k;
        for(int i=2;i<=n&&x>0;i++){
            k=(i==n?1:1ll<<(n-i-1));
            y=min(m-1,x/k);
            x-=y*k;
            c[i]+=y;
        }
        long long sum=s;
        a[1]=s;
        for(int i=2;i<=n;i++){
            a[i]=sum+1+c[i];
            sum+=a[i];
        }
        if(a[n]==t){
            printf("%d ",n);
            for(int i=1;i<=n;i++)printf("%lld ",a[i]);
            puts("");
        }
        else puts("-1");
    }
    return 0;
}

1166 E

The meaning of problems

solution