LightOJ 1006 - Hex-a-bonacci
2016-09-06 17:18
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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\n", ++caseno, fn(n) % 10000007);
}
return 0;
}
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.
PROBLEM SETTER: JANE ALAM JAN
很明显不能直接按题上的代码直接敲,题意代码就是一个递归的过程,把这个过程模拟出来即可,(记得当时做时,一直想着有什么规律。。。。)
PDF (English) | Statistics | Forum |
Time Limit: 0.5 second(s) | Memory Limit: 32 MB |
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
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 |
PROBLEM SETTER: JANE ALAM JAN
很明显不能直接按题上的代码直接敲,题意代码就是一个递归的过程,把这个过程模拟出来即可,(记得当时做时,一直想着有什么规律。。。。)
#include<stdio.h> #include<string.h> #include<algorithm> using namespace std; int g[11000]; int main() { int n,i,j,k,l,t,temp; scanf("%d",&t); k=1; while(t--) { memset(g,0,sizeof(g)); scanf("%d%d%d%d%d%d%d",&g[0],&g[1],&g[2],&g[3],&g[4],&g[5],&n); for(i=6;i<=n;i++) { temp=0; for(j=i-1;j>=i-6;j--) { temp+=g[j]%10000007; } g[i]=temp; } printf("Case %d: %d\n",k++,g %10000007); } return 0; }
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