Number Sequence

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 178534    Accepted Submission(s): 44354 Problem Description A number sequence is defined as follows:
f(1) = 1, f(2) = 1, f(n) = (A * f(n - 1) + B * f(n - 2)) mod 7.
Given A, B, and n, you are to calculate the value of f(n).
Input The input consists of multiple test cases. Each test case contains 3 integers A, B and n on a single line (1 <= A, B <= 1000, 1 <= n <= 100,000,000). Three zeros signal the end of input and this test case is not to be processed.
Output For each test case, print the value of f(n) on a single line.
Sample Input 1 1 3 1 2 10 0 0 0 Sample Output 2 5 一看這題立馬寫了個遞迴然後出了個這個Memory Limit Exceeded，，，記憶體超了，，

#include <stdio.h>
#include <iostream>
#include <algorithm>
#include <stdlib.h>
#include <string.h>

using namespace std;

int f(int A, int B, int n){
    if(n==1) return 1;
    if(n==2) return 1;
    return (A*f(A, B, n-1)+B*f(A, B, n-2))%7;
}
int main(){

    int n, A, B;
    while(scanf("%d%d%d",&A,&B,&n), A&&B&&n){
        printf("%d\n",f(A,B,n));
    }

    return 0;
}
 

然後發現應該是個迴圈，再寫被T了；

#include <stdio.h>
#include <iostream>
#include <algorithm>
#include <stdlib.h>
#include <string.h>

using namespace std;

int main(){

    int n, A, B;
    while(scanf("%d%d%d",&A,&B,&n)){
        if(A==0 && B==0 && n==0) break;
        int ff[1000];
        ff[1]=ff[2]=1;
        int i=3;
        for(i=3; i<=n; i++){
            ff[i]=(A*ff[i-1]+B*ff[i-2])%7;
           // printf("%d ",ff[i]);
            if(ff[i]==1&&ff[i-1]==1) break;
        }
        //printf("\n");
        if(i>n) printf("%d\n", ff[n]);
        else{
        int cnt=i-2;
        int ans;
        if(n%cnt==0) ans=cnt;
        else ans=n%cnt;
        printf("%d\n",ff[ans]);
        }
    }

    return 0;
}
 

然後再改，RE，，，，

#include <stdio.h>
#include <iostream>
#include <algorithm>
#include <stdlib.h>
#include <string.h>

using namespace std;

int main(){

    int n, A, B;
    while(scanf("%d%d%d",&A,&B,&n)){
        if(A==0 && B==0 && n==0) break;
        int ff[100];
        ff[1]=ff[2]=1;
        int i=3;
        for(i=3; i<=n; i++){
            ff[i]=(A*ff[i-1]+B*ff[i-2])%7;
            //printf("%d ",ff[i]);
            if(ff[i]==1&&ff[i-1]==1) break;
        }
       // printf("\n");
        if(i>n) printf("%d\n", ff[n]);
        else{
        int cnt=i-2;
       // printf("cnt: %d\n",cnt);
        ff[0]=ff[cnt];
        printf("%d\n",ff[n%cnt]);
        }
    }

    return 0;
}

當我把for迴圈裡的n改成50就A了，改成100還是RE，醉了；

#include <stdio.h>
#include <iostream>
#include <algorithm>
#include <stdlib.h>
#include <string.h>

using namespace std;

int main(){

    int n, A, B;
    while(scanf("%d%d%d",&A,&B,&n)){
        if(A==0 && B==0 && n==0) break;
        int ff[100];
        ff[1]=ff[2]=1;
        int i=3;
        for(i=3; i<=50; i++){        //就是這裡的50，改成n就RE，，，，，，，
            ff[i]=(A*ff[i-1]+B*ff[i-2])%7;
            //printf("%d ",ff[i]);
            if(ff[i]==1&&ff[i-1]==1) break;
        }
       // printf("\n");
        if(i>n) printf("%d\n", ff[n]);
        else{
        int cnt=i-2;
       // printf("cnt: %d\n",cnt);
        ff[0]=ff[cnt];
        printf("%d\n",ff[n%cnt]);
        }
    }

    return 0;
}

最後附上大神的AC程式碼，就是強啊，咋想到的呢，，，，，

#include <iostream>

using namespace std;

int main()
{
    int A,B,n,i;
    int a[48];
    cin >> A >> B >> n;
    while( A!=0 || B!=0 || n!=0)
    {
        a[0]=a[1]=1;
        for(i=2;i<48;i++)
            a[i] = ( A * a[i-1] + B * a[i-2]) % 7;
        cout << a[(n-1)%48] <<endl;//48個數必定迴圈一遍，或以48的因子數一個迴圈......膜了； 
        cin >> A >> B >> n;
    }
    return 0;
}