three

A

题意:就是你最多可以交换一次某两个元素, 尽可能的使得第i个位置是i的元素尽可能多, 问最多有多少个

思路:交换一次至少可以增加一个, 是增加一个还是两个判断一下就好

#include <cstdio>

const int qq = 1e5 + 10;
int num[qq];

int main(){
	int n; scanf("%d", &n);
	int cnt = 0;
	for(int i = 0; i < n; ++i){
		scanf("%d", num + i);
		if(num[i] == i){
			cnt++;
		}
	}
	for(int i = 0; i < n; ++i){
        if(num[i] == i) continue;
		if(num[num[i]] == i){
			printf("%d\n", cnt + 2);
			return 0;
		}
	}
	if (cnt != n) {
		printf("%d\n", cnt + 1);
 	} else {
 		printf("%d\n", cnt);
 	}
}

B

题意:给出编号为1到3的原子的度数, 问 1-2 2-3 3-1 原子之间各有多少条线,

思路: 假设 1-2 , 2-3, 3-1,原子之间各有x, y, z根线, 则x + y = b, y + z = c, x + z = a, 那么我们可以枚举x, 根据a, b, c可以求出y, z, 判断y, z 是否合法就可以

#include <cstdio>
#include <cstdlib>
#include <cmath>
#include <algorithm>
using namespace std;
const int qq = 1005;
const int N = 1e9 - 500;
int num[qq];

int main(){
	int a, b, c; scanf("%d%d%d", &a, &b, &c);
	int x, y, z;
	for(int i = 0; i <= a; ++i){
		x = i;
		z = a - x;
		y = c - z;
		if(z < 0 || y < 0)	continue;
		if(x + y == b){
			printf("%d %d %d\n", x, y, z);
			return 0;
		}
	}
	puts("Impossible");
	return 0;
}

C

随便求一下就好

#include <cstdio>

const int qq = 1e5 + 10;
int num[qq];

int main(){
	int n; scanf("%d", &n);
	int cnt = 0;
	int pre;
	int x;
	for(int i = 0; i < n; ++i){
        if(i == 0){
            scanf("%d", &pre);
            cnt++;
            continue;
        }
        scanf("%d", &x);
        if(x == pre)	continue;
        pre = x;
        cnt++;
	}
	printf("%d\n", cnt);
}

D

题意:你可以使得k个0 变成 1, 使得连续的1最多

思路: two point  维护每个合法区间, k = 0的情况直接特判掉了 暂时还没想到普遍性的解法

#include <cstdio>
#include <cstdlib>
#include <cmath>
#include <algorithm>
using namespace std;
const int qq = 3e5 + 10;
int num[qq];

int main(){
	int n, k; scanf("%d%d", &n, &k);
	for(int i = 0; i < n; ++i)	scanf("%d", num + i);
	int maxn = -1;
	int l, r, res, id;
	l = r = res = 0;
	if(k == 0){
		for(int i = 0; i < n; ++i){
			if(num[i] == 1)	res++;
			else	res = 0;
			maxn = max(res, maxn);
		}
		printf("%d\n", maxn);
		for(int i = 0; i < n; ++i)
			printf("%d ", num[i]);
		puts("");
		return 0;
	}
	while(r < n){
		if(num[r] == 0){
			if (k) {
				k--;
			} else {
				while(num[l] != 0) l++;
				l++;
			}
 		}
		if(maxn < r - l + 1){
			maxn = r - l + 1;
			id = l;
		}
		r++;
	}
	printf("%d\n", maxn);
	for(int i = 0; i < n; ++i){
		if (i >= id && i <= id + maxn -1) {
			printf("%d ", 1);
		} else {
			printf("%d ", num[i]);
		}
	}
	puts("");
	return 0;
}

E

题意:你需要在两个gcd != 1 的数之间加一个数, 使得两个相邻数之间的gcd = 1

思路 : 找出接近1e9 的最大的三个素数, 然后按规则加就好

#include <cstdio>
#include <cstdlib>
#include <cmath>
#include <algorithm>
using namespace std;
const int qq = 1005;
const int N = 1e9 - 500;
int num[qq];
int prime[5] = {999999893, 999999929, 999999937};

int gcd(int x, int y){
	return y == 0 ? x : gcd(y, x % y);
}

int main(){
	/*for(int i = N ; i <= 1000000000; ++i){
		bool flag = true;
		for(int j = 2; j <= sqrt(i) + 1; ++j)
			if(i % j == 0){
				flag = false;
				break;
			}
		if (flag) {
			printf("%d\n", i);
		}
	}*/
	int n; scanf("%d", &n);
	for(int i = 0; i < n; ++i)	scanf("%d", num + i);
	for(int i = 1; i < n; ++i){
		if (gcd(num[i-1], num[i]) == 1) {
			printf("%d ", num[i-1]);
		} else {
			printf("%d ", num[i-1]);
			for(int j = 0; j < 3; ++j){
				if(num[i-1] != prime[j] && num[i] != prime[j]){
					printf("%d ", prime[j]);
					break;
				}
			}
		}
	}
	printf("%d\n", num[n-1]);
	return 0;
}