codeforces round #424 div2

A 暴力查询，分三段查就可以了

#include<bits/stdc++.h>
using namespace std;
const int N = 110;
int n, pos;
int a[N];
int main()
{
    scanf("%d", &n);
    for(int i = 1; i <= n; ++i) scanf("%d", &a[i]);
    pos = -1;
    for(int i = 2; i <= n; ++i) if(a[i] - a[i - 1] <= 0)
    {
        pos = i;
        break;
    }
    if(pos == -1)
    {
        puts("YES");
        return 0;
    }
    for(int i = pos; i <= n; ++i) 
    {
        pos = 0;
        if(a[i] - a[i - 1] > 0)
        {
            puts("NO");
            return 0;
        }
        else if(a[i] - a[i - 1] < 0)
        {
            pos = i;
            break;
        }
    }
    if(!pos)
    {
        puts("YES");
        return 0; 
    }
    for(int i = pos; i <= n; ++i)
    {
        if(a[i] - a[i - 1] >= 0)
        {
            puts("NO");
            return 0;
        }
    }
    puts("YES");
    return 0;
}

B 开个数组映射一下

#include<bits/stdc++.h>
using namespace std;
const int N = 1010;
int n;
char s1[N], s2[N], s[N];
int Map[N];
int main()
{
    scanf("%s%s%s", s1 + 1, s2 + 1, s + 1);
    n = strlen(s1 + 1);
    for(int i = 1; i <= n; ++i) Map[s1[i] - 'a'] = s2[i] - 'a';
    for(int i = 1; i <= strlen(s + 1); ++i)
    {
        if(isdigit(s[i]))     
            printf("%c", s[i]);
        if(islower(s[i]))
            printf("%c", (char)(Map[s[i] - 'a'] + 'a'));
        if(isupper(s[i]))
            printf("%c", (char)(Map[s[i] - 'A'] + 'A'));    
    }    
    return 0;
}

C 没想出来，这种题目发现用b在a上算不行就应该转换用a在b上算。统计每个b对于每个a产生的初始值，如果一个初始值出现了k次，那么就是可以的。

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const int N = 2010;
ll tot;
int n, k, ans;
ll a[N];
map<ll, int> mp;
vector<ll> v;
inline int read()
{
    int x = 0, f = 1; char c = getchar();
    while(c < '0' || c > '9') { if(c == '-') f = -1; c = getchar(); }
    while(c >= '0' && c <= '9') { x = x * 10 + c - '0'; c = getchar(); }
    return x * f;
}
int main()
{
    scanf("%d%d", &n, &k);
    for(int i = 1; i <= n; ++i) 
    { 
        a[i] = read() + a[i - 1];
        v.push_back(a[i]);
    } 
    sort(v.begin(), v.end());
    v.erase(unique(v.begin(), v.end()), v.end());
    for(register int i = 1; i <= k; ++i)
    {
        int x;
        x = read();
        for(int j = 0; j < v.size(); ++j)
            ++mp[x - v[j]];
    }
    for(map<ll, int> :: iterator it = mp.begin(); it != mp.end(); ++it)
        ans += (it -> second == k);    
    printf("%d
", ans);
    return 0;
}

D dp，dp[i][j]表示第i个人选到第j个钥匙，dp[i][j]=min(dp[i][j-1],max(dp[i-1][j-1],abs(a[i]-b[j])+(b[j]-p))就好了

看错题目+C题 这道题1小时35分才出来。。。

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const int N = 2010;
int n, k, p;
int a[N], b[N];
ll dp[N][N];
int main()
{
    scanf("%d%d%d", &n, &k, &p);
    for(int i = 1; i <= n; ++i) scanf("%d", &a[i]);
    for(int i = 1; i <= k; ++i) scanf("%d", &b[i]);
    sort(a + 1, a + n + 1);
    sort(b + 1, b + k + 1);
    for(int i = 1; i <= n; ++i) dp[i][0]= 1e16;
    for(int i = 1; i <= n; ++i)
        for(int j = 1; j <= k; ++j)
            dp[i][j] = min(dp[i][j - 1], max(dp[i - 1][j - 1], (ll)abs(a[i] - b[j]) + (ll)abs(b[j] - p)));
    ll ans = 1e16;
    for(int i = n; i <= k; ++i)
        ans = min(ans, dp[n][i]);
    printf("%lld
", ans);
    return 0;
}

E 平衡树练习题 其实很简单就能搞定。我们发现如果选了一轮序列就会恢复原来的样子，只是删除了一些数。那么我们开set记录每个数出现的位置，从最小的数开始枚举，如果没到结尾加上上次到这次的距离，否则回到开头，因为每个数只删除一次，我们用size记录需要加上的答案，每删除一次size减1，到了一轮结束加上size。

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const int N = 100010;
int n, sz;
ll ans;
int a[N], tree[N];
set<int> s[N];
int main()
{
    scanf("%d", &n);
    for(int i = 1; i <= n; ++i)
    {
        scanf("%d", &a[i]);
        s[a[i]].insert(i);
    }
    sort(a + 1, a + n + 1);
    int last = 0;
    ans = n;
    sz = n;
    for(int i = 1; i <= n; ++i)
    {
        set<int> :: iterator it = s[a[i]].lower_bound(last);
        if(it == s[a[i]].end())
        {
            ans += sz;
            it = s[a[i]].begin();
        }
        last = *it;
        --sz;
        s[a[i]].erase(it);
    }
    printf("%lld
", ans);
    return 0;
}