CF798

链接：http://codeforces.com/contest/798

A:Mike and palindrome

 考虑不全面wa一次，水题。

#include<cstdio>
#include<cstring>
char s[20];

int main()
{
    scanf("%s",s);
    int n=strlen(s);
    int ct=0;
    int m=n/2;

    for(int i=0;i<m;i++)
    {
        if(s[i]!=s[n-i-1]) ct++;
        if(ct>1)
        {
            puts("NO");
            return 0;
        }
    }
    if(ct==0&&n%2==0)
    {
        puts("NO");
        return 0;
    }
    puts("YES");
    return 0;

}

B:Mike and strings

数据小，我暴力过的。。。

#include<cstdio>
#include<cstring>
char s[50][120];


int main()
{
    int n;
    scanf("%d",&n);
    for(int i=0;i<n;i++)
        scanf("%s",s[i]);
    int len=strlen(s[0]);
    int ans=0x3f3f3f3f;
    for(int i=0;i<n;i++)
    {
        int temp=0;
        for(int j=0;j<n;j++) if(i!=j)
        {
            for(int k=len;k<=2*len;k++)
                s[j][k]=s[j][k-len];
            int d=strstr(s[j],s[i])-s[j];
            if(d<0||d>50) {
                puts("-1");
                return 0;
            }
            s[j][len]='';  //
            temp+=d;
        }
        ans=ans>temp?temp:ans;
    }
    printf("%d
",ans);

}

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

# include <bits/stdc++.h>
using namespace std;
# define fi cin
# define fo cout
string s[512];
int dp[512][512];
int main(void)
{
    int n;
    fi>>n;
    for (int i = 1;i <= n;++i)
        fi>>s[i];
    int k = s[1].length();
    for (int i = 1;i <= n;++i)
        for (int j = 0;j < k;++j)
            dp[i][j] = 1e9;
    for (int i = 1;i <= n;++i)
        s[i] = s[i] + s[i];
    for (int i = 0;i < k;++i)
        dp[1][i] = i;
    for (int i = 2;i <= n;++i)
    {
        for (int j = 0;j < k;++j)
            for (int prev = 0;prev < k;++prev)
                if (s[i].substr(j,k) == s[i-1].substr(prev,k))
                    dp[i][j] = min(dp[i][j],dp[i-1][prev] + j);
    }
    int ans = 1e9;
    for (int i = 0;i < k;++i)
        ans = min(ans,dp[n][i]);
    if (ans == 1e9)
        puts("-1");
    else
        fo << ans << '
';
    return 0;
}

 C:Mike and gcd problem

有点巧妙，想不出

代码也很简洁

# include <bits/stdc++.h>
using namespace std;
# define fi cin
# define fo cout
int main(void)
{
    int n;
    fi>>n;
    int g = 0,v,cnt = 0,ans = 0;
    while (n --)
    {
        int v;
        fi>>v;
        g = __gcd(g,v);
        if (v & 1) ++cnt;
        else ans += (cnt / 2) + 2 * (cnt & 1),cnt = 0;
    }
    ans += (cnt / 2) + 2 * (cnt & 1);
    fo << "YES
";
    if (g == 1)
        fo << ans << '
';
    else
        fo << "0
";
    cerr << "Time elapsed :" << clock() * 1000.0 / CLOCKS_PER_SEC << " ms" << '
';
    return 0;
}

 D：Mike and distribution

慢慢要学会自己分析-_-||

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<vector>
using namespace std;
const int maxn=1e5+10;
#define pii pair<int,int>
#define se second
#define fi first
#define pb push_back

pii a[maxn];
int b[maxn];

int main()
{
    int n,x;
    scanf("%d",&n);
    for(int i=1;i<=n;i++)
    {
        scanf("%d",&x);
        a[i]=pii(-x,i);  //为了从大到小排序(-x,i)
    }
    sort(a+1,a+n+1);
    for(int i=1;i<=n;i++)
        scanf("%d",&b[i]);
    vector<int> v;
    v.pb(a[1].se);
    for(int i=2;i<=n;i+=2)
    {
        if(i+1<=n&&b[a[i+1].se]>b[a[i].se])
            v.pb(a[i+1].se);
        else v.pb(a[i].se);
    }
    int len=v.size();
    printf("%d
",len);
    for(int i=0;i<len-1;i++)
        printf("%d ",v[i]);
    printf("%d
",v[len-1]);
}

 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。。。