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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\n", ++caseno, fn(n) % 10000007);
}
return 0;
}
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.
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\n", t, fn(n) % mod); } return 0; }
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原文地址：http://www.cnblogs.com/pk28/p/5573400.html
