2020-2021:teams:manespace:codeforces_round_656_div._3
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2020-2021:teams:manespace:codeforces_round_656_div._3 [2020/07/24 11:18] intouchables 创建 |
2020-2021:teams:manespace:codeforces_round_656_div._3 [2020/07/24 11:43] (当前版本) intouchables [C] |
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| 比赛链接：https://codeforc.es/contest/1385 | 比赛链接：https://codeforc.es/contest/1385 | ||
| - | pro：3.5/7 | + | pro：3.5 / 7 |
| - | ====A==== | + | =====A===== |
| + | *题意：多组数据，每组三个正整数$x, y, z$，构造$a, b, c$满足$x = max(a, b)$, $y = max(a, c)$, $z= max(b, c)$ | ||
| + | *题解：签到题，分类特判即可 | ||
| + | <hidden ac代码> | ||
| + | <code c++> | ||
| + | #include<bits/stdc++.h> | ||
| + | using namespace std; | ||
| + | typedef long long ll; | ||
| + | const int maxn = 2e5 + 5; | ||
| + | const double pi = acos(-1); | ||
| + | const int mod = 998244353; | ||
| - | ====B==== | + | int x, y, z, a, b, c; |
| + | int t; | ||
| + | int main(){ | ||
| + | cin >> t; | ||
| + | while(t--){ | ||
| + | cin >> x >> y >> z; | ||
| + | int a = (x == y) + (x == z) + (y == z); | ||
| + | if(!a){ | ||
| + | puts("NO"); | ||
| + | continue; | ||
| + | } | ||
| + | else if(a == 3){ | ||
| + | puts("YES"); | ||
| + | cout << x << ' ' << x << ' ' << x << endl; | ||
| + | continue; | ||
| + | } | ||
| + | else{ | ||
| + | if((x == y && z > x) || (x == z && y > x) || (y == z && x > y)) puts("NO"); | ||
| + | else{ | ||
| + | puts("YES"); | ||
| + | if(x == y) cout << x << ' ' << z << ' ' << z << endl; | ||
| + | else if(x == z) cout << x << ' ' << y << ' ' << y << endl; | ||
| + | else cout << y << ' ' << x << ' ' << x << endl; | ||
| + | } | ||
| + | } | ||
| + | } | ||
| + | } | ||
| + | </code> | ||
| + | </hidden> | ||
| + | =====B===== | ||
| - | ====C==== | + | *题意：给定多组有且仅有两次重复的数字序列，在不影响相对顺序的前提下去重 |
| + | *题解：签到题，打标记跳输出即可 | ||
| + | <hidden ac代码> | ||
| + | <code c++> | ||
| + | #include<bits/stdc++.h> | ||
| + | using namespace std; | ||
| + | typedef long long ll; | ||
| + | const int maxn = 2e5 + 5; | ||
| + | const double pi = acos(-1); | ||
| + | const int mod = 998244353; | ||
| + | int t; | ||
| + | int n; | ||
| + | int a[105]; | ||
| - | ====D==== | + | int main(){ |
| + | cin >> t; | ||
| + | while(t--){ | ||
| + | cin >> n; | ||
| + | int tmp; | ||
| + | memset(a, 0, sizeof(a)); | ||
| + | for(int i = 1; i <= 2 * n; ++i){ | ||
| + | cin >> tmp; | ||
| + | if(a[tmp]) continue; | ||
| + | cout << tmp << ' '; | ||
| + | a[tmp] = 1; | ||
| + | } | ||
| + | puts(""); | ||
| + | } | ||
| + | } | ||
| + | </code> | ||
| + | </hidden> | ||
| - | 见本周推荐 | + | =====C===== |
| + | |||
| + | *题意：定义“good”序列：每次从首位或末位拿出元素，组成的新序列单调不降。多组数据，给定序列，询问删去从首位开始的至少多少元素，可以使序列成为“good”序列？ | ||
| + | *题解：所谓“good”序列就是从首末开始到某一位置都单调不降的序列，只需倒序遍历，先找不降序列，由此后找不升序列，统计剩余元素个数输出即可 | ||
| + | <hidden ac代码> | ||
| + | <code c++> | ||
| + | #include<bits/stdc++.h> | ||
| + | using namespace std; | ||
| + | |||
| + | typedef long long ll; | ||
| + | const int maxn = 2e5 + 5; | ||
| + | const double pi = acos(-1); | ||
| + | const int mod = 998244353; | ||
| + | |||
| + | int t; | ||
| + | int n; | ||
| + | int a[maxn]; | ||
| + | |||
| + | int main(){ | ||
| + | cin >> t; | ||
| + | while(t--){ | ||
| + | cin >> n; | ||
| + | for(int i = 1; i <= n; ++i) cin >> a[i]; | ||
| + | if(n == 1){ | ||
| + | cout << 0 << endl; | ||
| + | continue; | ||
| + | } | ||
| + | int p = -1, q = -1; | ||
| + | for(int i = n; i > 1; --i){ | ||
| + | if(a[i-1] < a[i]){ | ||
| + | p = i; | ||
| + | break; | ||
| + | } | ||
| + | } | ||
| + | if(p == -1){ | ||
| + | cout << 0 << endl; | ||
| + | continue; | ||
| + | } | ||
| + | for(int i = p; i > 1; --i){ | ||
| + | if(a[i-1] > a[i]){ | ||
| + | q = i; | ||
| + | break; | ||
| + | } | ||
| + | } | ||
| + | if(q == -1) cout << 0 << endl; | ||
| + | else cout << q - 1 << endl; | ||
| + | } | ||
| + | } | ||
| + | </code> | ||
| + | </hidden> | ||
| + | |||
| + | =====D===== | ||
| + | |||
| + | *见本周推荐 | ||
2020-2021/teams/manespace/codeforces_round_656_div._3.1595560732.txt.gz · 最后更改: 2020/07/24 11:18 由 intouchables
