Educational Codeforces Round 92 (Rated for Div. 2) A、B、C
2020-07-30 18:33
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目录
A. LCM Problem
题意
- 链接:LCM Problem
- 给出 l 、r ,求 x 、y 满足 l<=x<y<=r 且 l<=LCM(x,y)<=r
解题思路
- LCM(x,y)至少是min(x,y)的2倍,所以直接构造一组最小的解(l,2l),若2l>r则无解
代码
#include<stdio.h> #include<string.h> #include<algorithm> using namespace std; int main() { int t,l,r; scanf("%d",&t); while(t--) { scanf("%d%d",&l,&r); if(2*l<=r)printf("%d %d\n",l,2*l); else printf("-1 -1\n"); } return 0; }
B. Array Walk
题意
- 链接:Array Walk
- 给出一个长为n的数组a,一开始站在a1处得分a[i],现在规定你必须走k步,每步可以选择向左或向右一步取得相应的得分,但不能连续向左走且最多只能向左走z步,问最多能得多少分
解题思路
- 因为不能连续向左走,所以每次向左走完必定向右走,并且是在最大相邻之和的位置上选择向左走的,由于最多能向左走的步数很小,所以可以直接枚举向左左的步数
- sum[i]表示a[1]到a[i]的和
f[i]表示前i位最大相邻之和
代码
#include<stdio.h> #include<string.h> #include<algorithm> using namespace std; const int maxn=1e5+5; typedef long long ll; ll t,n,k,z,a[maxn],sum[maxn],f[maxn]; /* sum[i]表示a[1]到a[i]的和 f[i]表示前i位最大相邻之和 */ int main() { scanf("%d",&t); while(t--) { scanf("%lld%lld%lld",&n,&k,&z); sum[0]=0;f[0]=0;a[0]=0; for(ll i=1;i<=n;i++) { scanf("%lld",&a[i]); sum[i]=f[i]=0; sum[i]=sum[i-1]+a[i]; f[i]=max(f[i-1],a[i]+a[i-1]); } ll cnt=min(z,k/2),ans=sum[k+1];//限制左移次数 for(ll i=0;i<=cnt;i++)//左移i位 { ll ed=k+1-i*2;//最后的落点 ans=max(ans,sum[ed]+i*f[ed+1]); } printf("%lld\n",ans); } return 0; }
C. Good String
题意
- 链接:Good String
- 定义一种Good String为字符串t=t1t2t3…tn−1tn,它满足t2t3…tn−1tnt1= tnt1t2t3…tn−1
现给出一个仅有0-9组成的字符串s判断它最少要删除多少个字符才能变成Good String
解题思路
- 因为Good String满足 t2t3…tn−1tnt1= tnt1t2t3…tn−2tn−1 ,所以Good String要满足t[i]=t[i+2]即奇数位置上的数要相同且偶数位置上的数要相同
- 当Good String长度(即n)为奇数时,t
=t[2]所以整个字符串上的数都要相等;当Good String长度(即n)为偶数时,肯定由ABAB……AB组成 - 我们求出满足Good String的最长串的长度ans,再用原字符串s的长度len-ans就是答案
- 当最长串的长度ans为奇数或A=B时,肯定为原字符串s中出现次数最多的字符的次数,我们以这种情况初始化;然后因为A、B仅有0-9组成,我们直接枚举,判断每种情况下最长串的长度然后取最大即可
代码
#include<stdio.h> #include<string.h> #include<algorithm> using namespace std; const int maxn=2e5+5; typedef long long ll; char s[maxn]; int t,cnt[maxn]; int length(int x,int y,int len) { int sum=0,f=0; char xx=x+'0',yy=y+'0'; for(int i=0;i<len;i++) { if(f==0&&s[i]==xx)f=1; if(f==1&&s[i]==yy) { f=0;sum+=2; } } return sum; } int main() { scanf("%lld\n",&t); while(t--) { scanf("%s",s); int len=strlen(s),ans=-1; memset(cnt,0,sizeof(cnt)); for(int i=0;i<len;i++) { int tmp=s[i]-'0'; cnt[tmp]++; ans=max(ans,cnt[tmp]); } for(int i=0;i<=9;i++) for(int j=0;j<=9;j++) { if(i==j)continue; ans=max(ans,length(i,j,len)); } printf("%d\n",len-ans); } return 0; }
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