Educational Codeforces Round 93 (Rated for Div. 2)

目录

• Educational Codeforces Round 93 (Rated for Div. 2)
  • A. Bad Triangle
  • B. Substring Removal Game
  • C. Good Subarrays
  • D. Colored Rectangles
  • 参考

Educational Codeforces Round 93 (Rated for Div. 2)

A. Bad Triangle

题意： 给定长度为n的数组，给定n个数字，a[i] <= a[i + 1], 问是否可以找到ai,aj,ak，使得这三个数字组不成三角形。
题解： 一开始看错题意，找三角形，写了半个小时发现要找不能作三角形的。只要判断前两个数字和最后一个数字便可。由于前两个数字最小，最后一个数字最大。
代码：

#include <bits/stdc++.h>

using namespace std;

int const N = 1e5 + 10;
int a[N], n, T;

int main() {
    cin >> T;
    while (T--) {
        cin >> n;
        for (int i = 1; i <= n; ++i) scanf("%d", &a[i]);
        if (a[1] + a[2] <= a[n]) cout << 1 << " " << 2 << " " << n << endl;
        else cout << -1 << endl;
    }
    return 0;
}

B. Substring Removal Game

题意： 给你一个长度为n的0,1字符串，两我的轮流选择其中很多于一位的连续子序列，得分为所选子序列中1个数的总和，你先选，问最多可以得到多少分。
题解： 把连续的1分段拿出来，排序完倒叙，奇数和累加就是得分
代码：

#include <bits/stdc++.h>

using namespace std;

int const N = 1e5 + 10;
int a[N], n, T;

int main() {
    cin >> T;
    while (T--) {
        string s;
        cin >> s;
        vector<int> cnt;
        int l = 0, r = 0;
        while (s[r] == '0' && r < s.size()) r++;
        // cout << r << endl;
        l = r;
        // cout << l << " " << r << endl; 
        while (1) {
            while (s[r] == '1' && r < s.size()) r++;
            // cout << r << endl;
            cnt.push_back(r - l);
            while (s[r] == '0' && r < s.size()) r++;
            l = r;
            if (r == s.size()) break;
        }
        
        int res = 0;
        sort(cnt.begin(), cnt.end());
        reverse(cnt.begin(), cnt.end());
        for (int i = 0; i < cnt.size(); ++i) {
            if (i % 2 == 0) {
                res += cnt[i];
            }
        }
        cout << res << endl;
    }
    return 0;
}

C. Good Subarrays

题意： 给你T组数据，每组给你一个长度为n的字符串，求知足
\(\sum_{i=l}^{r}ai=r-l+1\) 的子区间的个数(i < j)
题解： 把\(\sum_{i=l}^{r}ai=r-l+1\)左右两边都减去r-l+1,则得\(\sum_{i=l}^{r}(ai-1)=0\)，因此只须要找区间和为0得个数。找区间和为0得个数能够使用前缀统计的方式，具体见代码。
代码：

#include <bits/stdc++.h>

using namespace std;

typedef long long LL;
int const N = 1e5 + 10;
LL T, n, a[N], sum[N];
string s;
map<LL, LL> mp;

int main() {
    cin >> T;
    while (T--) {
        cin >> n;
        cin >> s;
        mp.clear();
        for (int i = 0; i < s.size(); ++i) a[i + 1] = s[i] - '0' - 1;
        for (int i = 1; i <= n; ++i) sum[i] = sum[i - 1] + a[i];
        // for (int i = 1; i <= n; ++i) cout << a[i] << endl;
        LL res = 0;
        // int l = 1;
        mp[0]++;
        // while (a[l] == 0 && l < n) res ++, l ++;
        for (int i = 1; i <= n; ++i) {
            // if (a[i] == 0) res++;
            res += mp[sum[i]];
            mp[sum[i]]++;
        }
        cout << res << endl;
    }
}

// 1 2 0
// 0 1 -1
// 1 + 
// 0 0 0

// 11011
// 0 0 -1 0 0
// 1 + 1 + 1 + 1 +

D. Colored Rectangles

题意： 有R，G，B三种颜色的棍子。如今告诉你每种棍子长度的对数和每对的长度，让你用两种不一样颜色的棍子对围成矩形，求每一个矩形加起来的最大值
题解： 把每一个棍子拿起来从大到小排序，而后dp暴力转移
代码：

#include <bits/stdc++.h>

using namespace std;

typedef long long LL;
int const N = 210;
LL f[N][N][N], r, g, b, A[N], B[N], C[N];

int main() {
    cin >> r >> g >> b;
    memset(f, 0, sizeof f);
    for (int i = 1; i <= r; ++i) scanf("%lld", &A[i]);
    for (int i = 1; i <= g; ++i) scanf("%lld", &B[i]);
    for (int i = 1; i <= b; ++i) scanf("%lld", &C[i]);
    sort(A + 1, A + r + 1);
    reverse(A + 1, A + r + 1);
    sort(B + 1, B + g + 1);
    reverse(B + 1, B + g + 1);
    sort(C + 1, C + b + 1);
    reverse(C + 1, C + b + 1);
    for (int i = 0; i <= r; ++i) {
        for (int j = 0; j  <= g; ++j) {
            for (int k = 0; k <= b; ++k) {
                LL t1 = 0 , t2 = 0, t3 = 0;
                if (i && j)  t1 = f[i - 1][j - 1][k] + A[i] * B[j];
                if (i && k)  t2 = f[i - 1][j][k - 1] + A[i] * C[k];
                if (j && k)  t3 = f[i][j - 1][k - 1] + B[j] * C[k];
                
                f[i][j][k] = max(max(t1, f[i][j][k]), max(t2, t3));
            }
        }
    }
    LL res = 0;
    for (int i = 0; i <= r; ++i) 
        for (int j = 0; j  <= g; ++j) 
            for (int k = 0; k <= b; ++k) 
                res = max(res, f[i][j][k]);
    cout << res << endl;
    return 0;
}

参考

https://blog.csdn.net/xmyrzb/article/details/108020212