LightOJ - 1006 - Hex-a-bonacci


Statment

 #include<bits/stdc++.h>
using namespace std;
#define sf(a)                   scanf("%I64d",&a)
#define sff(a,b)                scanf("%I64d %I64d",&a,&b)
#define sfff(a,b,c)             scanf("%I64d %I64d %I64d",&a,&b,&c)
#define sffl(a,b,c)             scanf("%lld %lld %lld",&a,&b,&c)
#define sffff(a,b,c,d)          scanf("%I64d %I64d %I64d %I64d",&a,&b,&c,&d)
#define sfffff(a,b,c,d,e)       scanf("%I64d %I64d %I64d %I64d %I64d",&a,&b,&c,&d,&e)
#define read                    freopen("input.txt","r",stdin)
#define write                   freopen("output.txt","w",stdout)
#define ll                      long long
#define MAX                     1000009
#define Max                     100000000009
#define pb                      push_back
#define all(c)                  c.begin(),c.end()
#define mset(a,b)               memset(a,b,sizeof(a))
#define MOD                     10000007
#define ff                      first
#define sc                      second

 int main()
{
    ll dp[10009];
    vector<ll>a;
    ll c;
    scanf("%lld",&c);
    for(ll j=1; j<=c; j++)
    {
        a.clear();
        mset(dp,0);
        ll n,x;
        ll i;
        for(i=0; i<6; i++)
        {
            scanf("%lld",&x);
            a.pb(x);
        }
        scanf("%lld",&n);
        dp[0]=a[0]%MOD;
        for(i=1; i<6; i++)
            dp[i]=a[i]%MOD;
        for(i=6; i<=n; i++)
            dp[i]=(dp[i-1]%MOD+dp[i-2]%MOD+dp[i-3]%MOD+dp[i-4]%MOD+dp[i-5]%MOD+dp[i-6]%MOD)%MOD;
        printf("Case %lld: %lld\n",j,dp[n]);
    }
}

মন্তব্যসমূহ

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এই ব্লগটি থেকে জনপ্রিয় পোস্টগুলি

CP3 Exercise 1.2.3//5

#include<bits/stdc++.h> using namespace std; #define sf(a)                   scanf("%I64d",&a) #define sff(a,b)                scanf("%I64d %I64d",&a,&b) #define sfff(a,b,c)             scanf("%I64d %I64d %I64d",&a,&b,&c) #define sffff(a,b,c,d)          scanf("%I64d %I64d %I64d %I64d",&a,&b,&c,&d) #define sfffff(a,b,c,d,e)       scanf("%I64d %I64d %I64d %I64d %I64d",&a,&b,&c,&d,&e) #define flop(m,n,q)             for(ll i=m;i<n;i+=q) #define read                    freopen("input.txt","r",stdin) #define write                   freopen("output.txt","w",stdout) #define ll                   ...
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C++ Class

ছবি
> Abstraction >E ncapsulation   /Access Modifiers >Inheritance #include <iostream> using namespace std; class vehicle { public:     vehicle()     {         cout<<"this is a constructor of vehicle"<<endl;     }     ~vehicle()     {         cout<<"this is a deconstructor of vehicle"<<endl;     } }; class fourwheeler { public:     fourwheeler()     {         cout<<"this is the constructor of fourwheeler"<<endl;     }     fourwheeler(string x)     {         cout<<"This is another constructor which has "<<x<<" with it"<<endl;     }     ~fourwheeler()     {         cout<<"this is deconstructor of fourwheeler"<<endl;   ...
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SPOJ-INVERSE-SORT

Problem Algorithm: We have to generate  10!  combination of " abcdefghij  " by reversing and put appropriate distance from " abcdefghij  " So we need  key(string)->value  data structure. First thing comes to our mind is map<string,int> But it uses O(logN) time to insert a value as everything is ordered here So we use unordered_map ( Normally o(1) highest o(n) barely )  After that we take 2 string input and then find the relative string S from s1->s2 and another AC!! CODE Time needed 40-45sec .
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