Codeforces Round #446 (Div. 2)

A

#include <cstdio>#include <cstring>#include <cmath>#include <cstdlib>#include <ctime>#include <iostream>#include <algorithm>#include <sstream>#include <string>#include <vector>#include <queue>#include <stack>#include <map>#include <set>#include <utility>#include <bitset>using namespace std;#define LL long long#define pb push_back#define mk make_pair#define fi first#define se second#define lson (rt << 1)#define rson ((rt << 1) | 1)#define pill pair<int, int>#define mst(a, b)memset(a, b, sizeof a)#define REP(i, x, n)for(int i = x; i <= n; ++i)const int MOD = 1e9 + 7;const int qq = 1e5 + 10;const LL INF = 1e18 + 10;LL a[qq];int main(){int n;scanf("%d", &n);LL sum = 0;LL x;for(int i = 0; i < n; ++i) {scanf("%lld", &x);sum += x;}for(int i = 0; i < n; ++i) {scanf("%lld", a + i);}sort(a, a + n);if(sum > a[n - 1] + a[n - 2]) {puts("NO");} else {puts("YES");}return 0;}

B

倒着遍历维护每次能杀到最前面的id

#include <cstdio>#include <cstring>#include <cmath>#include <cstdlib>#include <ctime>#include <iostream>#include <algorithm>#include <sstream>#include <string>#include <vector>#include <queue>#include <stack>#include <map>#include <set>#include <utility>#include <bitset>using namespace std;#define LL long long#define pb push_back#define mk make_pair#define fi first#define se second#define lson (rt << 1)#define rson ((rt << 1) | 1)#define pill pair<int, int>#define mst(a, b)memset(a, b, sizeof a)#define REP(i, x, n)for(int i = x; i <= n; ++i)const int MOD = 1e9 + 7;const int qq = 1e6 + 10;const LL INF = 1e18 + 10;int num[qq];int main(){int n;scanf("%d", &n);for(int i = 1; i <= n; ++i) {scanf("%d", num + i);}int maxn = n + 1;int ans = n;for(int i = n; i >= 1; --i) {if(maxn <= i)ans--;maxn = min(maxn, i - num[i]);}printf("%d\n", ans);return 0;}

C

题意：每次可以求一下相邻两个数的gcd并用这gcd代替这两个数中的其中一个，问最后序列全部为1最少操作多少次，不可以就输出-1

思路：很明显如果求出整个序列的gcd > 1的话肯定是-1， 如果序列中存在1的情况的话只需要n - (1的个数)就是操作数了，剩下情况就n^2暴力枚举求出每一段的gcd，求出最小一段的gcd = 1的然后答案就是这个长度 + n - 1

#include <cstdio>#include <cstring>#include <cmath>#include <cstdlib>#include <ctime>#include <iostream>#include <algorithm>#include <sstream>#include <string>#include <vector>#include <queue>#include <stack>#include <map>#include <set>#include <utility>#include <bitset>using namespace std;#define LL long long#define pb push_back#define mk make_pair#define fi first#define se second#define lson (rt << 1)#define rson ((rt << 1) | 1)#define pill pair<int, int>#define mst(a, b)memset(a, b, sizeof a)#define REP(i, x, n)for(int i = x; i <= n; ++i)const int MOD = 1e9 + 7;const int qq = 1e5 + 10;const int INF = 1e9 + 10;int gcd(int a, int b) {return b == 0 ? a : gcd(b, a % b);}int num[2005];int main(){int n;scanf("%d", &n);bool f = false;int tmp;for(int i = 1; i <= n; ++i) {scanf("%d", num + i);if(num[i] == 1)f = true;if(i == 1)tmp = num[i];elsetmp = gcd(tmp, num[i]);}if(f) {int cnt = 0;for(int i = 1; i <= n; ++i) {if(num[i] == 1)cnt++;}printf("%d\n", n - cnt);return 0;}if(tmp > 1) {printf("-1\n");return 0;}int minx = INF;for(int j, i = 2; i <= n; ++i) {int x = num[i - 1];for(j = i; j <= n; ++j) {x = gcd(x, num[j]);if(x == 1)break;}if(x == 1) {minx = min(minx, j - (i - 1));}}printf("%d\n", minx + (n - 1));return 0;}

D

题意：给出序列a，将a序列重排一下成为b序列，使得a序列的所有非空子序列和不等于b序列的子序列和

思路：题目说了所有数都不相同，那么我们可以构造b序列中的数都比其对应a序列中的数恰好大一点点，

比如 1 2 3 4

         2 3 4 1

这里除了a序列中最大的4对应b序列的1意外其他的b序列中对应的都比a序列要大

我们可以这样分析，选出的子序列假设a中的和是x1， 那么a序列中未被选中的和是x2，b序列中被选中的子序列和是y1，为被选中的和是y2

可以知道 x1 + x2 = y1 + y2

如果x1中不包含a序列的最大值，那么可以知道 x1 < y1

如果x1中包含a序列的最大值，那么可以知道 x2 < y2 由最初的等式就知道 x1 > y1

#include<bits/stdc++.h>using namespace std;#define LL long long#define pb push_back#define fi first#define se second#define mk make_pair#define mst(a, b)memset(a, b, sizeof a)const int qq = 1e5 + 10;const int MOD = 1e9 + 7;const int INF = 1e9 + 10;int a[qq], b[qq], c[qq];map<int, int> mp, rk;int main() {int n;scanf("%d", &n);for(int i = 1; i <= n; ++i) {scanf("%d", a + i);c[i] = a[i];}sort(c + 1, c + 1 + n);for(int i = 1; i <= n; ++i) {rk[c[i]] = i;mp[i] = c[i];}for(int i = 1; i <= n; ++i) {int tmp = rk[a[i]];b[i] = mp[(tmp == n ? 1 : tmp + 1)];}for(int i = 1; i <= n; ++i) {printf("%d ", b[i]);}puts("");return 0;}