CF798

考虑不全面wa一次，水题。

 1 #include<cstdio>
 2 #include<cstring>
 3 char s[20];
 4 
 5 int main()
 6 {
 7     scanf("%s",s);
 8     int n=strlen(s);
 9     int ct=0;
10     int m=n/2;
11 
12     for(int i=0;i<m;i++)
13     {
14         if(s[i]!=s[n-i-1]) ct++;
15         if(ct>1)
16         {
17             puts("NO");
18             return 0;
19         }
20     }
21     if(ct==0&&n%2==0)
22     {
23         puts("NO");
24         return 0;
25     }
26     puts("YES");
27     return 0;
28 
29 }

View Code

B:Mike and strings

数据小，我暴力过的。。。

 1 #include<cstdio>
 2 #include<cstring>
 3 char s[50][120];
 4 
 5 
 6 int main()
 7 {
 8     int n;
 9     scanf("%d",&n);
10     for(int i=0;i<n;i++)
11         scanf("%s",s[i]);
12     int len=strlen(s[0]);
13     int ans=0x3f3f3f3f;
14     for(int i=0;i<n;i++)
15     {
16         int temp=0;
17         for(int j=0;j<n;j++) if(i!=j)
18         {
19             for(int k=len;k<=2*len;k++)
20                 s[j][k]=s[j][k-len];
21             int d=strstr(s[j],s[i])-s[j];
22             if(d<0||d>50) {
23                 puts("-1");
24                 return 0;
25             }
26             s[j][len]='';  //
27             temp+=d;
28         }
29         ans=ans>temp?temp:ans;
30     }
31     printf("%d
",ans);
32 
33 }

View

参考：

用dp做

dp[i][j]表示使得前i个串和第i个串移动j次之后都相同的最少移动的次数

 1 # include <bits/stdc++.h>
 2 using namespace std;
 3 # define fi cin
 4 # define fo cout
 5 string s[512];
 6 int dp[512][512];
 7 int main(void)
 8 {
 9     int n;
10     fi>>n;
11     for (int i = 1;i <= n;++i)
12         fi>>s[i];
13     int k = s[1].length();
14     for (int i = 1;i <= n;++i)
15         for (int j = 0;j < k;++j)
16             dp[i][j] = 1e9;
17     for (int i = 1;i <= n;++i)
18         s[i] = s[i] + s[i];
19     for (int i = 0;i < k;++i)
20         dp[1][i] = i;
21     for (int i = 2;i <= n;++i)
22     {
23         for (int j = 0;j < k;++j)
24             for (int prev = 0;prev < k;++prev)
25                 if (s[i].substr(j,k) == s[i-1].substr(prev,k))
26                     dp[i][j] = min(dp[i][j],dp[i-1][prev] + j);
27     }
28     int ans = 1e9;
29     for (int i = 0;i < k;++i)
30         ans = min(ans,dp[n][i]);
31     if (ans == 1e9)
32         puts("-1");
33     else
34         fo << ans << '
';
35     return 0;
36 }

View Code

C:Mike and gcd problem

有点巧妙，想不出

代码也很简洁

 1 # include <bits/stdc++.h>
 2 using namespace std;
 3 # define fi cin
 4 # define fo cout
 5 int main(void)
 6 {
 7     int n;
 8     fi>>n;
 9     int g = 0,v,cnt = 0,ans = 0;
10     while (n --)
11     {
12         int v;
13         fi>>v;
14         g = __gcd(g,v);
15         if (v & 1) ++cnt;
16         else ans += (cnt / 2) + 2 * (cnt & 1),cnt = 0;
17     }
18     ans += (cnt / 2) + 2 * (cnt & 1);
19     fo << "YES
";
20     if (g == 1)
21         fo << ans << '
';
22     else
23         fo << "0
";
24     cerr << "Time elapsed :" << clock() * 1000.0 / CLOCKS_PER_SEC << " ms" << '
';
25     return 0;
26 }

View Code

D：Mike and distribution

慢慢要学会自己分析-_-||

 1 #include<cstdio>
 2 #include<cstring>
 3 #include<algorithm>
 4 #include<vector>
 5 using namespace std;
 6 const int maxn=1e5+10;
 7 #define pii pair<int,int>
 8 #define se second
 9 #define fi first
10 #define pb push_back
11 
12 pii a[maxn];
13 int b[maxn];
14 
15 int main()
16 {
17     int n,x;
18     scanf("%d",&n);
19     for(int i=1;i<=n;i++)
20     {
21         scanf("%d",&x);
22         a[i]=pii(-x,i);  //为了从大到小排序(-x,i)
23     }
24     sort(a+1,a+n+1);
25     for(int i=1;i<=n;i++)
26         scanf("%d",&b[i]);
27     vector<int> v;
28     v.pb(a[1].se);
29     for(int i=2;i<=n;i+=2)
30     {
31         if(i+1<=n&&b[a[i+1].se]>b[a[i].se])
32             v.pb(a[i+1].se);
33         else v.pb(a[i].se);
34     }
35     int len=v.size();
36     printf("%d
",len);
37     for(int i=0;i<len-1;i++)
38         printf("%d ",v[i]);
39     printf("%d
",v[len-1]);
40 }

View Code

E:Mike and code of a permutation

貌似有点难，貌似用到线段树和拓扑排序。

等有中文题解了再补吧。。。