Educational Codeforces Round 86 (Rated for Div. 2)

题目链接：http://codeforces.com/contest/1342

A

思路：一个是一个数加一减一，花费a元，一个是两个数同时加一减一，花费b元，那么判断b和2a的大小即可，a肯定是乘以x和y的差值的，加上公共部分即min(x,y)乘以b还是乘以2a,加上他俩最小的即可.

//-------------------------------------------------
//Created by HanJinyu
//Created Time :日  4/26 22:29:40 2020
//File Name :86A.cpp
//-------------------------------------------------

#include <stdio.h>
#include <string.h>
#include <iostream>
#include <algorithm>
#include <vector>
#include <queue>
#include <set>
#include <map>
#include <list>
#include <map>
#include <string>
#include <math.h>
#include <stdlib.h>
#include <time.h>
using namespace std;
typedef double db;
typedef long long ll;
const int maxn = 200005;

int main()
{
      //freopen("in.txt","r",stdin);
      //freopen("out.txt","w",stdout);
      int t;
      scanf("%d",&t);
      while(t--)
      {

          ll x,y,a,b;
          scanf("%lld%lld%lld%lld",&x,&y,&a,&b);
          printf("%lld
",(max(x,y)-min(x,y))*a+min(min(x,y)*b,min(x,y)*2*a));
      }
 
     return 0;
}

B

思路：给你一个01字符串，要求找到一个字符使得他是所给字符串的子串，且长度<=2len，且该字符串的周期尽可能小，那么直接输出长度为2*n的01字符串即可：10101010...

当所给字符串只有一种字符时，输出原有字符串即可。

//-------------------------------------------------
//Created by HanJinyu
//Created Time :日  4/26 23:25:55 2020
//File Name :86B.cpp
//-------------------------------------------------

#include <stdio.h>
#include <string.h>
#include <iostream>
#include <algorithm>
#include <vector>
#include <queue>
#include <set>
#include <map>
#include <list>
#include <map>
#include <string>
#include <math.h>
#include <stdlib.h>
#include <time.h>
using namespace std;
typedef double db;
typedef long long ll;
const int maxn = 200005;

int main()
{
     // freopen("in.txt","r",stdin);
    //freopen("out.txt","w",stdout);
    int T;
    scanf("%d",&T);
    while(T--)
    {

        char str[200];
        scanf("%s",str);
        int len=strlen(str);
        int _0=0,_1=0;
        for(int i=0;i<len;i++)
        {
            if(str[i]=='0')
                _0++;
            else
                _1++;
        }
        if(_0==0||_1==0)
            printf("%s
",str);
        else
        {
            for(int i=0;i<len;i++)
                printf("10");
           // printf("1
");
           printf("
");
        }
    }
 
     return 0;
}

C

思路：打表样例中的1-200就行了，将i%a%b==i%b%a的数全部列举出来即可，就会发现有规律，即每个a*b里面的满足条件的个数都是一样的，这样就可以用前缀和来写，满足条件的置1，否则为0，再求前缀和，这样所得的就是该区间满足条件的个数了，通过%(a*b)来计算个数就好

//-------------------------------------------------
//Created by HanJinyu
//Created Time :一  4/27 00:09:16 2020
//File Name :86C.cpp
//-------------------------------------------------

#include <stdio.h>
#include <string.h>
#include <iostream>
#include <algorithm>
#include <vector>
#include <queue>
#include <set>
#include <map>
#include <list>
#include <map>
#include <string>
#include <math.h>
#include <stdlib.h>
#include <time.h>
using namespace std;
typedef double db;
typedef long long ll;
const int maxn = 200005;

int main()
{
   // freopen("in.txt","r",stdin);
    //freopen("out.txt","w",stdout);
    int T;
    scanf("%d",&T);
    while(T--)
    {

        int a,b,k;
        scanf("%d%d%d",&a,&b,&k);
        int sum[maxn];
          for(int i=1;i<=a*b;i++)
            {

                if(i%a%b!=i%b%a)
                    sum[i]=1;
                else
                    sum[i]=0;
                sum[i]+=sum[i-1];
            }

        while(k--)
        {

            ll l,r;
            scanf("%lld%lld",&l,&r);
            ll ans=(r-l)/(a*b)*sum[a*b];
            ll lll=l%(a*b);
            ll rrr=r%(a*b);
            if(rrr>=lll)
                printf("%lld ",ans+sum[rrr]-sum[lll-1]);
            else
                printf("%lld ",ans+sum[rrr]+sum[a*b]-sum[lll-1]);            
        }
                    printf("
");

    }
 
     return 0;
}