A. Array Divisibility
思路:
找出特例,发现输出 1∼𝑛 符合题意。直接输出1~n即可.
代码:
#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
#define N 1000005
ll dp[N], w[N], v[N], h[N];
ll dis[1005][1005];
ll a, b, c, n, m, t;
ll ans, sum, num, minn = 1e9 + 7, maxx = 0;
struct node {
ll a, b, c;
}f[N];
string s1, s2;
int main()
{
cin >> t;
while (t--)
{
cin >> n;
for (int i = 1; i <= n; i++)
{
cout << i << " ";
}
cout << endl;
}
return 0;
}B. Corner Twist
思路:
关键的充要条件是 𝑎,𝑏 的每一行/列的和模 3 后相等。证明的话，首先要想到 2×2 的操作是可以完成所有大小的子矩阵操作的，手模一下可以发现是对的。接着考虑比较暴力的方式，我们遍历 𝑖,𝑗 从 1∼𝑛−1 和 1∼𝑚−1，然后把 𝑎𝑖,𝑗 按对应操作修改成 𝑏𝑖,𝑗，经过这样之后前 𝑛−1 行和 𝑚−1 列都是可以相等的，所以也就是看最后那一列是否满足。而这个遍历顺序又是可以变化的，也就是最后剩下的行列可以是任意一行一列，所以要所有行列均满足模 3 结果相等。
代码:
#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
#define N 1000005
ll dp[N], w[N], v[N], h[N];
char a[505][505], b[505][505];
ll x1[505], x2[505], y[505], y2[505];
ll n, m, t;
ll ans, sum, num, minn = 1e9 + 7, maxx = 0;
struct node {
ll a, b, c;
}f[N];
string s1, s2;
int main()
{
cin >> t;
while (t--)
{
cin >> n >> m;
memset(x1, 0, sizeof(x1));
memset(x2, 0, sizeof(x2));
memset(y, 0, sizeof(y));
memset(y2, 0, sizeof(y2));
for (int i = 1; i <= n; i++) {
for (int j = 1; j <= m; j++) {
cin >> a[i][j];
x1[i] += a[i][j]-'0';
y[j] += a[i][j]-'0';
}
}
for (int i = 1; i <= n; i++) {
for (int j = 1; j <= m; j++) {
cin >> b[i][j];
x2[i] += b[i][j]-'0';
y2[j] += b[i][j]-'0';
}
}
int flag = 1;
for (int i = 1; i <= n; i++) {
if (x1[i] % 3 != x2[i] % 3) {
flag = 0;
break;
}
}
for (int j = 1; j <= m; j++) {
if (y[j] % 3 != y2[j] % 3) {
flag = 0;
break;
}
}
if (flag == 1)
cout << "YES" << endl;
else
cout << "NO" << endl;
}
return 0;
}C. Have Your Cake and Eat It Too
思路:
枚举三个人的顺序，然后把序列分成三段就行了，判断可以直接 𝑂(𝑛)，至于枚举顺序不需要重复粘贴六次代码，用一个 next_permutation 函数就行了。
代码:
#include<bits/stdc++.h>
using namespace std;
#define N 1000005
typedef long long ll;
typedef unsigned long long ull;
ll n, m, t, h, k;
ll ans, num, sum, cnt;
ll dp[N], ac[4][N], a[4][N], get1[N], get2[N];
bool flag, vis[N];
struct node {
ll mark, x, y;
}f[N];
bool cmp(node a, node b) {
return a.mark < b.mark;
}
int main()
{
cin >> t;
while (t--){
cin >> n;
ac[1][0] = ac[2][0] = ac[3][0] = sum = 0;
for (int i = 1; i <= 3; i++) {
for (int j = 1; j <= n; j++) {
cin >> a[i][j];
if(i==1) sum += a[i][j];
ac[i][j] = ac[i][j - 1] + a[i][j];
}
}
sum = (sum + 2) / 3;
for (int i = 1; i <= 3; i++) {
for (int j = 1; j <= n; j++) {
if (ac[i][j] >= sum) {
get1[i] = j;
break;
}
}
}
for (int i = 1; i <= 3; i++) {
for (int j = n - 1; j >= 0; j--) {
if (ac[i][n]-ac[i][j] >= sum) {
get2[i] = j+1;
break;
}
}
}
ll order[3] = { 1,2,3 };
flag = 1;
do {
for (int i = 0; i < 3; i++)
f[i].mark = order[i];
f[1].x = get1[order[0]] + 1, f[1].y = get2[order[2]] - 1;
if (ac[order[1]][f[1].y] - ac[order[1]][f[1].x - 1] < sum)
continue;
f[0].x = 1, f[0].y = get1[order[0]];
f[2].x = get2[order[2]], f[2].y = n;
sort(f , f + 3, cmp);
for (int i = 0; i < 3; i++) {
cout << f[i].x << " " << f[i].y << " ";
}
cout << endl;
flag = 0;
break;
} while (next_permutation(order, order + 3));
if (flag)
cout << -1 << endl;
}
return 0;
}D. Swap Dilemma
思路:
首先如果元素不同直接输出"NO",如果相同,那么我们可以通过求出个两个序列的转换成相同序列的次数,也就是会改变 𝑎,𝑏 逆序对的数量，并且同时改变其数量奇偶性,所以怎么只要保证两次序列的奇偶性相同即可,但我们可以进行优化,将只对一个序列进行操作,只要将序列B映射到序列A上,那么问题就变成了只需要求映射后的序列B的逆序对个数即可.
代码:
#include<bits/stdc++.h>
using namespace std;
#define N 1000005
typedef long long ll;
typedef unsigned long long ull;
ll n, m, t, h, k;
ll ans, num, sum, cnt;
ll temp[N], a[N], f1[N], f2[N];
bool flag, vis[N];
struct node {
ll mark, x, y;
}f[N];
bool cmp(node a, node b) {
return a.mark < b.mark;
}
void merge_sort(int l, int r)
{
if (l == r)return;
int mid = l + r >> 1;
merge_sort(l, mid);
merge_sort(mid + 1, r);
int ll = l, st = l, rr = mid + 1;
while (l <= mid and rr <= r)
{
if (a[l] <= a[rr])
temp[ll++] = a[l++];
else
temp[ll++] = a[rr++], ans += mid - l + 1;
}
while (l <= mid)temp[ll++] = a[l++];
while (rr <= r)temp[ll++] = a[rr++];
for (int i = st; i <= r; i++)
a[i] = temp[i];
return;
}
int main()
{
cin >> t;
while (t--){
ans = 0,flag=1;
cin >> n;
map<ll, ll>mp;
for (int i = 1; i <= n; i++) {
cin >> f1[i];
mp[f1[i]] = i;
}
for (int i = 1; i <= n; i++) {
cin >> f2[i];
}
for (int i = 1; i <= n; i++) {
if (mp.count(f2[i]) == 0) {
flag = 0;
break;
}
a[i] = mp[f2[i]];
}
merge_sort(1, n);
if (flag == 0) {
cout << "NO" << endl;
}
else {
if (ans % 2==0)
cout << "YES" << endl;
else
cout << "NO" << endl;
}
}
return 0;
}
