CCPC长春赛重现

比赛链接
http://acm.hdu.edu.cn/contests/contest_show.php?cid=728

QUALITY的题解
https://async.icpc-camp.org/d/582-2016

02

分数加法除法，对着公式模拟即可

#include <iostream>
#include <cstdio>
using namespace std;
typedef long long LL;
LL gcd (LL a,LL b){
    return b==0?a:gcd(b,a%b);
}
struct fenshu{
    LL fz,fm;
    fenshu (){fz = 0;fm = 1;}
    fenshu (LL x,LL y):fz(x),fm(y){}
    fenshu operator + (const fenshu &a)const{
        LL FM = fm*a.fm/gcd(fm,a.fm) ;
        LL FZ = fz*FM/fm+a.fz*FM/a.fm;
        return fenshu{FZ,FM};
    }
    fenshu operator / (const fenshu &a)const{
        LL FZ = fz*a.fm;
        LL FM = fm*a.fz;
        LL gc = gcd(FZ,FM);
        return fenshu{FZ/gc,FM/gc};
    }
}a[11],b[11];

int main(){
    //freopen("in.txt","r",stdin);
    LL T; scanf("%I64d",&T);
    for (LL n,cas=1;cas<=T;cas++){
        scanf("%I64d",&n);
        for (LL i=1;i<=n;i++) a[i].fz=1 ,b[i].fz=1;
        for (LL i=1;i<=n;i++) scanf("%I64d",&a[i].fz);
        for (LL i=1;i<=n;i++) scanf("%I64d",&b[i].fz);
        fenshu ans = fenshu(0,1);
        for (LL i=n;i>=1;i--){
            ans = b[i]/(ans + a[i]);
        }
        printf("Case #%I64d: %I64d %I64d
",cas,ans.fz,ans.fm);
    }
    return 0;
}

04

#include <stdio.h>
#include <iostream>

using namespace std;
int main(){
    int a[21]={0,0,0,0,1,1,2,3,3,4,5,6,7,7,8,9,10,11,12,13,14},T,t=1;
    cin >> T;
    while (T--){
        int n;
        cin >> n;
        cout << "Case #" << t++ << ": " << a[n] << endl;
    }
}

06

#include <stdio.h>
#include <iostream>
#include <cstring>

using namespace std;
int main(){
    int a[10010]={},T,t=1;
    cin >> T;
    while (T--){
        int n,k,x=0;
        cin >> n >> k;
        memset(a,0,sizeof a);
        cout << "Case #" << t++ << ": ";
        for (int i=1;i<=k;i++) {
            printf("%d ",2*i);
            a[2*i]=1;
            x++;
        }
        for (int i=1;i<=n;i++){
            if (!a[i]){
                x++; printf("%d",i);
                if (x!=n) printf(" ");
            }
        }
        cout << endl;
    }
}

08
KMP，
next数组要改成Next数组，不然会CE
还有每次要memset一下a，b数组的初值

#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;
const int N = 1000100;
int a[N],b[N],Next[N];

void getNext(int m){
    int i=0,j=-1;
    Next[0] = -1;
    for (;i<m;) {
        if (j==-1||b[i]==b[j]){
            if (b[++i]!=b[++j])Next[i]=j;
            else Next[i] = Next[j];
        }else j = Next[j];
    }
}
int KMP(int start,int n,int m,int p){
    int ans=0,i=start,j=0;
    for (;i<n&&j<m;){
        if (j==-1||a[i]==b[j])i+=p,j++;
        else j = Next[j];
        if (!i&&!j)break;
        if (j==m){
            ans++;
            j = Next[j];
        }
    }
    return ans;
}

int main(){
    //freopen("in.txt","r",stdin);
    int T,n,m,p;scanf("%d",&T);
    for (int cas=1;cas<=T;cas++){
        scanf("%d%d%d",&n,&m,&p);
        memset(Next,0,sizeof(Next));
        memset(a,0,sizeof(a));
        memset(b,0,sizeof(b));
        for (int i=0;i<n;i++)scanf("%d",&a[i]);
        for (int i=0;i<m;i++)scanf("%d",&b[i]);
        int ans = 0;
        getNext(m);
        for (int i=0;i<p;i++) ans += KMP(i,n, m, p);
        printf("Case #%d: %d
",cas,ans);
    }
    return 0;
}

10
每次取前半段的数字-1，然后构造回文数，
每次可以减掉一半的位数，1000位很快就减完了
涉及高精度减法，偷懒写java了，
诶，java不熟，这辈子没写过这么丑的代码

import java.math.*;
//import java.io.*;
import java.util.*;
public class Main {
    public static BigInteger fanzhuan(BigInteger n){
        //System.out.println("fanzhuan" + n);
        int k = String.valueOf(n).length();
        BigInteger ret = BigInteger.ZERO;
        for (int i=1;i<=k;i++){
            ret = ret.add(n.mod(BigInteger.TEN));
            ret = ret.multiply(BigInteger.TEN);
            n = n.divide(BigInteger.TEN);
        }
        ret = ret.divide(BigInteger.TEN);
        return ret;
    }
    public static void main(String[] argv){
        Scanner cin = new Scanner(System.in);
        int T =cin.nextInt();
        for (int cas=1;cas<=T;cas++){
            BigInteger n = cin.nextBigInteger();
            int N = String.valueOf(n).length();
            int A = 0;
            BigInteger ans[] = new BigInteger[55];
            for (;N>1;){
                //System.out.println("n=" + n);
                BigInteger one0 = BigInteger.TEN.pow(N>>1);
                BigInteger half = n.divide(one0);
                if (N<=2){
                    if (N==1){ 
                        ans[++A] = n; 
                        n = BigInteger.ZERO;
                        break;
                    }else {//2 wei
                        if (n.compareTo(BigInteger.valueOf(19))==0){
                            ans[++A] = BigInteger.valueOf(11); 
                            ans[++A] = BigInteger.valueOf( 8); 
                            n = BigInteger.ZERO;
                            break;
                        }else if (n.compareTo(BigInteger.valueOf(19))==-1){
                            ans[++A] = BigInteger.valueOf(9);
                            ans[++A] = n.subtract(BigInteger.valueOf(9));
                            n = BigInteger.ZERO;
                            break;
                        }//else continue;
                    }
                }
                half = half.subtract(BigInteger.ONE);
                //System.out.println("half=" + half);
                BigInteger fan = fanzhuan(half);
                if (N%2>0) fan = fan.mod(one0);
                //System.out.println("fan=" + fan);
                BigInteger jan = half.multiply(one0).add(fan);
                //System.out.println("jan=" + jan);
                ans[++A] = jan ;
                n = n.subtract(jan);
                N = String.valueOf(n).length();
            }
            if (n.compareTo(BigInteger.ZERO)==1)ans[++A] = n;
            System.out.printf("Case #%d:
%d
",cas,A);
            for (int i=1;i<=A;i++){
                System.out.println(ans[i]);
            }
        }
        cin.close();
    }
}

黑大帅5题铜，不知道大连的时候我们也没有这个好运气
rp++