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  • Lightoj 1006 Hex-a-bonacci

    Given a code (not optimized), and necessary inputs, you have to find the output of the code for the inputs. The code is as follows:

    int a, b, c, d, e, f;
    int fn( int n ) {
        if( n == 0 ) return a;
        if( n == 1 ) return b;
        if( n == 2 ) return c;
        if( n == 3 ) return d;
        if( n == 4 ) return e;
        if( n == 5 ) return f;
        return( fn(n-1) + fn(n-2) + fn(n-3) + fn(n-4) + fn(n-5) + fn(n-6) );
    }
    int main() {
        int n, caseno = 0, cases;
        scanf("%d", &cases);
        while( cases-- ) {
            scanf("%d %d %d %d %d %d %d", &a, &b, &c, &d, &e, &f, &n);
            printf("Case %d: %d ", ++caseno, fn(n) % 10000007);
        }
        return 0;
    }

    Input

    Input starts with an integer T (≤ 100), denoting the number of test cases.

    Each case contains seven integers, a, b, c, d, e, f and n. All integers will be non-negative and 0 ≤ n ≤ 10000 and the each of the others will be fit into a 32-bit integer.

    Output

    For each case, print the output of the given code. The given code may have integer overflow problem in the compiler, so be careful.

    Sample Input

    Output for Sample Input

    5

    0 1 2 3 4 5 20

    3 2 1 5 0 1 9

    4 12 9 4 5 6 15

    9 8 7 6 5 4 3

    3 4 3 2 54 5 4

    Case 1: 216339

    Case 2: 79

    Case 3: 16636

    Case 4: 6

    Case 5: 54

    /* ***********************************************
    Author        :guanjun
    Created Time  :2016/6/10 10:54:05
    File Name     :1006.cpp
    ************************************************ */
    #include <iostream>
    #include <cstring>
    #include <cstdlib>
    #include <stdio.h>
    #include <algorithm>
    #include <vector>
    #include <queue>
    #include <set>
    #include <map>
    #include <string>
    #include <math.h>
    #include <stdlib.h>
    #include <iomanip>
    #include <list>
    #include <deque>
    #include <stack>
    #define ull unsigned long long
    #define ll long long
    #define mod 10000007
    #define INF 0x3f3f3f3f
    #define maxn 10010
    #define cle(a) memset(a,0,sizeof(a))
    const ull inf = 1LL << 61;
    const double eps=1e-5;
    using namespace std;
    priority_queue<int,vector<int>,greater<int> >pq;
    struct Node{
        int x,y;
    };
    struct cmp{
        bool operator()(Node a,Node b){
            if(a.x==b.x) return a.y> b.y;
            return a.x>b.x;
        }
    };
    
    bool cmp(int a,int b){
        return a>b;
    }
    ll dp[maxn];
    ll a, b, c, d, e, f,n;
    ll fn( int n ) {
        if(dp[n]!=-1)return dp[n];
        if( n == 0 ) return a;
        if( n == 1 ) return b;
        if( n == 2 ) return c;
        if( n == 3 ) return d;
        if( n == 4 ) return e;
        if( n == 5 ) return f;
        return dp[n]=(fn(n-1) + fn(n-2) + fn(n-3) + fn(n-4) + fn(n-5) + fn(n-6))%mod;
    }
    int main()
    {
        #ifndef ONLINE_JUDGE
        freopen("in.txt","r",stdin);
        #endif
        //freopen("out.txt","w",stdout);
        int T;
        cin>>T;
        for(int t=1;t<=T;t++){
            memset(dp,-1,sizeof dp);
            cin>>a>>b>>c>>d>>e>>f>>n;
            printf("Case %d: %lld
    ", t, fn(n) % mod);
        }
        return 0;
    }
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  • 原文地址:https://www.cnblogs.com/pk28/p/5573400.html
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