Codeforces Round 909 (Div. 3)
https://codeforces.com/contest/1899
一个小游戏
#include<iostream> #include<cstdio> #include<cstring> #include<algorithm> using namespace std; int n; int main() { cin >> n; while(n -- ) { int x; cin >> x; if((x + 1) % 3 == 0 || (x - 1) % 3 == 0) { cout << "First" << endl; } else { cout << "Second" << endl; } } return 0; }
将数组分成k分，使绝对值差值最大
#include<iostream> #include<cstdio> #include<cstring> #include<algorithm> #include<set> using namespace std; const int N = 150010; typedef long long LL; int n; int a[N]; LL s[N]; int main() { int T; scanf("%d", &T); while(T -- ) { memset(s, 0, sizeof s); scanf("%d", &n); for(int i = 1; i <= n; i ++ ) scanf("%d", &a[i]); for(int i = 1; i <= n; i ++ ) s[i] = s[i - 1] + a[i]; LL res = 0; for(int k = 1; k <= n / 2; k ++ ) { if(n % k == 0) { LL mi = 1e18, mx = 0; for(int j = k; j <= n; j += k) { mi = min(mi, s[j] - s[j - k]); mx = max(mx, s[j] - s[j - k]); } res = max(res, mx - mi); } } cout << res << endl; } return 0; }
找一个和最大的子数组，要求相邻的元素不能有相同的奇偶性
简单动态规划问题，另f[i]表示以第i项为结尾的，满足条件的最大子序列和。
#include<iostream> #include<cstdio> #include<cstring> #include<algorithm> using namespace std; const int N = 2e5 + 10; int n; int a[N], f[N]; int main() { int T; scanf("%d", &T); while(T -- ) { scanf("%d", &n); for(int i = 0; i < n; i ++ ) scanf("%d", &a[i]), f[i] = a[i]; for(int i = 1; i < n; i ++ ) { if(((a[i] & 1) == (a[i - 1] & 1))) continue; f[i] = max(f[i], f[i - 1] + a[i]); } int mx = -1e8; for(int i = 0; i < n; i ++ ) mx = max(mx, f[i]); printf("%d\n", mx); } return 0; }
推公式，按照题目给的条件进行化简即可。
#include<iostream> #include<cstdio> #include<cstring> #include<algorithm> #include<map> #define LL long long using namespace std; const int N = 2e5 + 10; inline int read(){ int r = 0,w = 1; char c = getchar(); while (c < '0' || c > '9'){ if (c == '-') w = -1; c = getchar(); } while (c >= '0' && c <= '9'){ r = (r << 3) + (r << 1) + (c ^ 48); c = getchar(); } return r * w; } int main() { int n; int T = read(); LL res; int x; map<int, int> mp; map<int,int>::iterator it; while(T -- ) { res = 0; scanf("%d", &n); mp.clear(); for(register int i = 0; i < n; ++ i) { x = read(); ++ mp[x]; } // 注意这里不能次次声明，会被卡住 for(it = mp.begin(); it != mp.end(); ++it) { res += it -> second * 1ll * (it -> second - 1) / 2; } res += mp[1] * 1ll * mp[2]; printf("%lld\n", res); } return 0; }
贪心，找规律。
我们要得到一个非降序的的序列，操作二可以把小的前移，直到大于等于前一个元素。若我们一直执行操作二，最后会得到一个类似于样例1的序列，然后进行操作一，将大元素插到末尾即可。
在此过程中，如果最小值前面有山峰状数组及形如 3 4 2 的子序列，则无解。
#include<iostream> #include<cstdio> #include<cstring> #include<algorithm> #include<climits> using namespace std; const int N = 2e5 + 10; int main() { int T; scanf("%d", &T); int a[N]; int n; while(T -- ) { scanf("%d", &n); for(int i = 0; i < n; i ++ ) scanf("%d", &a[i]); int mi = INT_MAX, p = -1; for(int i = 0; i < n; i ++ ) { if(a[i] < mi) { mi = a[i]; p = i; } } bool flag = true; for(int i = p; i < n - 1; i ++ ) { if(a[i] > a[i + 1]) { flag = false; break; } } if(flag) printf("%d\n", p); else printf("%d\n", -1); } return 0; }
https://codeforces.com/contest/1899/problem/F
贪心，维护一条链，让链尾部最后一个节点进行移动即可满足题目条件。
#include<iostream> #include<cstdio> #include<cstring> #include<algorithm> using namespace std; const int N = 510; int n, q; int a[N]; int main() { int T; scanf("%d", &T); int len; while(T -- ) { scanf("%d%d", &n, &q); len = n - 1; for(int i = 0; i < q; i ++ ) { scanf("%d", &a[i]); } for(int i = 1; i < n; i ++ ) { printf("%d %d\n", i, i + 1); } int f = n - 1; for(int i = 0; i < q; i ++ ) { if(a[i] == f) { printf("%d %d %d\n", -1, -1, -1); } else { printf("%d %d %d\n", n, f, a[i]); f = a[i]; } } } return 0; }
最后一题太难，贴个链接有空再说：https://codeforces.com/contest/1899/problem/G
