Educational Codeforces Round 46 (Rated for Div. 2)

A - Codehorses T-shirts

思路：有相同抵消，没有相同的对答案+1

#include<bits/stdc++.h>
#define LL long long
#define fi first
#define se second
#define mk make_pair
#define pii pair<int, int>

using namespace std;

const int N = 1e5 + ;
const int M = 1e5 + ;
const int inf = 0x3f3f3f3f;
const LL INF = 0x3f3f3f3f3f3f3f3f;
const int mod = 1e9 + ;

string a[], b[];
int n;

map<string, int> mp;
int main(){
    scanf("%d", &n);
    for(int i = ; i <= n; i++) cin >> a[i], mp[a[i]]++;
    int ans = ;
    for(int i = ; i <= n; i++) {
        cin >> b[i];
        if(mp[b[i]]) {
            mp[b[i]]--;
        } else {
            ans++;
        }
    }
    printf("%d\n", ans);
    return ;
}

/*
*/

B - Light It Up

思路：求一下前缀，如果要加一个开关肯定是加在原来是关的前面一个或者后面一个，枚举一下就好啦。

#include<bits/stdc++.h>
#define LL long long
#define fi first
#define se second
#define mk make_pair
#define pii pair<int, int>

using namespace std;

const int N = 1e5 + ;
const int inf = 0x3f3f3f3f;
const LL INF = 0x3f3f3f3f3f3f3f3f;
const int mod = ;

int n, M, sum[N], fsum[N], a[N];
int main(){
    scanf("%d%d", &n, &M);
    for(int i = ; i <= n; i++) scanf("%d", &a[i]);
    a[++n] = M;
    for(int i = ; i <= n; i++) {
        sum[i] = sum[i - ];
        if(i & ) sum[i] += a[i] - a[i - ];
        fsum[i] = a[i] - sum[i];
    }

    int ans = sum[n];
    for(int i = ; i < n; i++) {
        if(i & ) {
            if(a[i] > a[i - ] + ) {
                int ret = a[i] - a[i - ] - ;
                ret += fsum[n] - fsum[i];
                ret += sum[i - ];
                ans = max(ans, ret);
            }

            if(a[i] < a[i + ] - ) {
                int ret = sum[i];
                ret += fsum[n] - fsum[i + ];
                ret += a[i + ] - a[i] - ;
                ans = max(ans, ret);
            }
        }
    }

    printf("%d\n", ans);
    return ;
}

/*
*/

C - Covered Points Count

思路：离散化差分标记搞一搞

#include<bits/stdc++.h>
#define LL long long
#define fi first
#define se second
#define mk make_pair
#define pii pair<int, int>

using namespace std;

const int N = 1e6 +  + 2e5;
const int M = 1e5 + ;
const int inf = 0x3f3f3f3f;
const LL INF = 0x3f3f3f3f3f3f3f3f;
const int mod = 1e9 + ;

LL n, tot, in[N], out[N], ans[N];
LL hs[N], l[N], r[N];

int main(){
    scanf("%lld", &n);
    for(int i = ; i <= n; i++) {
        scanf("%lld%lld", &l[i], &r[i]);
        hs[++tot] = l[i];
        hs[++tot] = r[i];
        hs[++tot] = l[i] + ;
        hs[++tot] = l[i] - ;
        hs[++tot] = r[i] + ;
        hs[++tot] = r[i] - ;
    }

    sort(hs + , hs +  + tot);
    tot = unique(hs + , hs +  + tot) - hs - ;

    for(int i = ; i <= n; i++) {
        int pos1 = lower_bound(hs + , hs +  + tot, l[i]) - hs;
        int pos2 = lower_bound(hs + , hs +  + tot, r[i]) - hs;
        in[pos1]++;
        out[pos2]++;
    }

    LL cnt = ;
    for(int i = ; i < tot; i++) {
        cnt += in[i];
        ans[cnt]++;
        cnt -= out[i];
        LL num = hs[i + ] - hs[i] - ;
        ans[cnt] += num;
    }

    cnt += in[tot];
    ans[cnt]++;

    for(int i = ; i <= n; i++) printf("%lld ", ans[i]);
    return ;
}

/*
*/

D - Yet Another Problem On a Subsequence

题目大意：给你n个数字 (n <= 1000)， 问你有多少个子序列是good  sequence

good sequence的定义是能变成若干个good arrays

good array 的定义是这个数组的第一个元素的值等于这个数组的length - 1。

思路: A了三题之后日常开始挂机。。。 还是太菜啦！！！！

比赛的时候感觉是个dp，但是不知到如何定义状态。。。

后来发现原来我看错了题，原来是把good sequence划分成若干个good arrays

我们定义dp[ i ]，表示从第 i 个开始，并且 i 作为一个array第一个元素的的合法答案的方案数。

#include<bits/stdc++.h>
#define LL long long
#define fi first
#define se second
#define mk make_pair
#define pii pair<int, int>

using namespace std;

const int N = 1e3 + ;
const int M = 1e5 + ;
const int inf = 0x3f3f3f3f;
const LL INF = 0x3f3f3f3f3f3f3f3f;
const int mod = ;

int a[N], dp[N], C[N][N], n;

void init() {
    for(int i = (C[][] = ); i < N; i++)
        for(int j = (C[i][] = ); j < N; j++)
            C[i][j] = (C[i - ][j] + C[i - ][j - ]) % mod;
}

int main() {
    init();
    scanf("%d", &n);
    for(int i = ; i <= n; i++) {
        scanf("%d", &a[i]);
    }

    dp[n + ] = ;

    for(int i = n; i >= ; i--) {
        if(a[i] > ) {
            for(int j = i + a[i] + ; j <= n + ; j++) {
                dp[i] = (dp[i] + 1LL * dp[j] * C[j - i - ][a[i]]) % mod;
            }
        }
    }

    int ans = ;

    for(int i = ; i <= n; i++) {
        ans = (ans + dp[i]) % mod;
    }

    printf("%d\n", ans);
    return ;
}
/*
*/