codeforces--283--

这场比以往的难点额= =

A题就是 每次删掉2-N-1这(N-2)位置上的一个元素 然后观察 max(a[i]-a[i-1])。

#include <iostream>
#include <algorithm>
using namespace std;

int a[110];

int main()
{
    int n , ans , sum;
    sum = 11111111;
    ans = 0;
    cin >> n;
    cin >> a[1];
    for( int i = 2 ; i<=n ; i++ )
    {
        cin >> a[i];
        ans = max( ans , a[i]-a[i-1] );
    }
    for( int i = 2 ; i<=n-1 ; i++ )
    {
        sum = min( sum , a[i+1] - a[i-1] );
    }
    cout << max(sum,ans) << endl;
    return 0;
}

B题 add操作最多执行10次 每一次操作完成之后 再对整个字符串进行 循环移位

#include <iostream>
#include <string>
using namespace std;

string ans , str;

void solve( )
{
    string s , t;
    char ch;
    int len = str.length();
    int temp;
    for( int T = 1 ; T<=10 ; T++ )
    {
        s = "";
        for( int i = 0 ; i<len ; i++ )
        {
            temp = (str[i]-'0'+T)%10;//char转int
            ch = temp + '0';//int转char
            s += ch;
        }
        if( s<ans )
            ans = s;
        for( int j = 1 ; j<len ; j++ )
        {
            t = "";
            t.append(s,j,len-j);
            t.append(s,0,j);
            if( t<ans )
                ans = t;
        }
    }
}

int main()
{
    int n;
    cin >> n;
    cin >> str;
    ans = str;
    solve( );  
    cout << ans << endl;
}

C题 每次只能删除一列上的所有元素 字典序的排列是对行进行要求的 所以暴力遍历过去就好 注意标记 

#include <iostream>
#include <cstring>
using namespace std;

int n , m , cnt;
const int size = 110;
char str[size][size];
bool can[size];

void solve()
{
    bool flag;
    cnt = 0;
    memset( can , false , sizeof(can) );
    for( int i = 0 ; i<m ; i++ )
    {
        flag = true;
        for( int j = 0 ; j<n-1 ; j++ )
        {
            if( !can[j] && str[j][i] > str[j+1][i] )
            {
                ++ cnt;
                flag = false;
                break;
            }
        }
        if( flag )
        {
            for( int j = 0 ; j<n-1 ; j++ )
            {
                if( str[j][i] < str[j+1][i] )
                {
                    can[j] = true;    
                }
            }
        }
    }
}

int main()
{
    cin >> n >> m;
    for( int i = 0 ; i<n ; i++ )
    {
        cin >> str[i];
    }
    solve();
    cout << cnt << endl;
    return 0;
}

D题  porker说是个大模拟 我不会= =

#include <iostream>
#include <algorithm>
#include <cstring>
using namespace std;

class data {
public:
    int s, t;
    bool operator<(const data& p) const {
        if (s < p.s) return true;
        else if (s == p.s && t < p.t) return true;
        else return false;
    }
};

int sum[2][100100];
data ans[100100];
int counts;

int main() {
    int n;
    cin.sync_with_stdio(false);
    cin >> n;
    sum[0][0] = sum[0][1] = 0;
    for (int i = 1; i <= n; i++) {
        int d;
        cin >> d;
        d--;
        sum[0][i] = sum[0][i - 1];
        sum[1][i] = sum[1][i - 1];
        sum[d][i]++;
    }
    counts = 0;
    int maximum = max(sum[0][n], sum[1][n]);
    for (int i = 1; i <= maximum; i++) {
        int sets[2] = { 0, 0 };
        int index = 0;
        int interval = i * 2;
        bool flag = true;
        int lastwin = -1;
        while (index < n) {
            int nextindex = index + interval;
            int maxindex = index + interval;
            if (maxindex > n + 1) {
                maxindex = n + 1;
            }
            int first_index = lower_bound(sum[0] + index + 1, sum[0] + maxindex, sum[0][index] + i) - sum[0];
            int second_index = lower_bound(sum[1] + index + 1, sum[1] + maxindex, sum[1][index] + i) - sum[1];
            if (first_index < second_index) {
                sets[0]++;
                lastwin = 0;
            }
            else if (second_index < first_index) {
                sets[1]++;
                lastwin = 1;
            }
            else {
                flag = false;
            }
            index = min(first_index, second_index);
        }
        if (!flag) continue;
        if (sets[0] > sets[1] && lastwin == 0 || sets[0] < sets[1] && lastwin == 1) {
            ans[counts].s = max(sets[0], sets[1]);
            ans[counts].t = i;
            counts++;
        }
    }
    sort(ans, ans + counts);
    cout << counts << endl;
    for (int i = 0; i < counts; i++) {
        cout << ans[i].s << " " << ans[i].t << endl;
    }
    //    system("pause");
}