Guangdong University of Technology 2015 freshman Race Round2 (//own brain hole blocked, madan! ）

Path: Freshman Race

A.number Sequence

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 is to calculate the value of f (n).

Analyse:

Auto-find cycle method.

CODE:

int main () {    int a,b,n;    Cir[1]=cir[2]=1;    while (Cin>>a>>b>>n)    {        if (!a&&!b&&!n) break;        int num=0,i,j;        for (i=3;; i++)        {            cir[i]= (a*cir[i-1]+b*cir[i-2])%7;            cout<<cir[i]<< "--" <<cir[i-1]<<endl;            for (j=2;j<i;j++)            {                if (Cir[i-1]==cir[j-1]&&cir[i]==cir[j])                {                    num=i-j;                    break;                }            }            if (num) break;        }        if (n<=j)            printf ("%d\n", Cir[n]);        else            printf ("%d\n", cir[j+ (n-j)%num? ( N-J) (%num:num)]);    }    return 0;}

C. A shameless person

Description:

Give a long long number n, and let you ask for the sum.

Law One: | N | Modulo 10 method, The pit is the smallest negative number is-(1<<31) and the positive maximum is only (1<<31)-1 .... Need a special sentence

<span style= "color: #3333FF;" >n==-9223372036854775808</span>

Method Two: The positive solution is a string read. Well, this should not be a pit jump!!

while (Cin>>n)    {        LL ans=0;        if (n==-9223372036854775808) {                   //FA a pit            printf ("89\n");            Continue;        }        if (n<0) n=-n;        while (n)        {            ans+=n%10;            n/=10;        }        printf ("%lld\n", ans);    }

D. Coprimes

Description:

Give the number n, ask for less than N, and with the number of N, coprime.

Analyse:

Direct violence for the loop is good.

CODE:

int gcd (int a,int b) {    return b==0?a:gcd (b,a%b);}

E. Catch that Cow

Description:

Give the axis two points n,k, ask from N to K, each min can move from X to X-1 or x+1 or 2*x

Analyse:

Originally want to find the law directly, but still have small trick, oneself always did not find.

Positive solution: 0<=n<=100000,2^20, so the BFS direct search, but also only dozens of layers. Brain hole can't jam, madan!.

CODE:

typedef long LONG Ll;const int n=100007;int dp[2*n];void BFS (int n,int m) {mem (dp,-1);    Queue<int> que;    dp[n]=0;    Que.push (n);        while (!que.empty ()) {int C=que.front ();        Que.pop ();        cout<<c<< "-" <<dp[c]<<endl;        if (c==m) break;                if (c<m) {if (dp[c+1]==-1) {dp[c+1]=dp[c]+1;            Que.push (c+1);                } if (c&&dp[c-1]==-1) {dp[c-1]=dp[c]+1;            Que.push (c-1);                } if (c<m&&dp[2*c]==-1) {dp[2*c]=dp[c]+1;            Que.push (2*C);                }} else {if (dp[c-1]==-1) {dp[c-1]=dp[c]+1;            Que.push (c-1);    }}}}int Main () {int n,m;        while (scanf ("%d%d", &n,&m) ==2) {BFS (n,m);    printf ("%d\n", Dp[m]); } return 0;}

Guangdong University of Technology 2015 freshman Race Round2 (//own brain hole blocked, madan! ）