A. Array Divisibility
题目大意
构造长度为n的数组，满足：
-
所有j的aj之和可以被k整除，其中j是k的倍数，k的取值为1~n。
思路
构造序列1->n即可满足条件。
代码实现
void solve() {
ll n; cin >> n;
for (int i = 1; i <= n; i++)cout << i << " ";
cout << "\n";
}B. Corner Twist
题目大意
给定a和b两个网格，网格中只存在0、1、2三种值。
给定操作，可以选择一个矩形的一条对角线上的两个端点，让他们加1/2；另一条对角线的端点值加2/1，然后对这些值取模3。问是否存在操作可以让a和b两个网格一致。
思路
考虑网格的边上的值，边上的值的变化会受到其他的影响，所以我们优先去找如何让边上的值相等即可。
正扫一遍，反扫一遍，然后比较数组即可。
代码实现
void solve() {
ll n, m; cin >> n >> m;
vector<vector<char>>a(n + 1, vector<char>(m + 1));
vector<vector<char>>b(n + 1, vector<char>(m + 1));
for (int i = 1; i <= n; i++)for (int j = 1; j <= m; j++)cin >> a[i][j];
for (int i = 1; i <= n; i++)for (int j = 1; j <= m; j++)cin >> b[i][j];
for (int i = 1; i < n; i++) {
for (int j = 1; j < m; j++) {
if (a[i][j] != b[i][j]) {
if (a[i][j] == '0') {
if (b[i][j] == '1') {
a[i][j] += 1;
if (a[i][j] == '3')a[i][j] = '0';
if (a[i][j] == '4')a[i][j] = '1';
a[i + 1][j + 1] += 1;
if (a[i + 1][j + 1] == '3')a[i + 1][j + 1] = '0';
if (a[i + 1][j + 1] == '4')a[i + 1][j + 1] = '1';
a[i + 1][j] += 2;
if (a[i + 1][j] == '3')a[i + 1][j] = '0';
if (a[i + 1][j] == '4')a[i + 1][j] = '1';
a[i][j + 1] += 2;
if (a[i][j + 1] == '3')a[i][j + 1] = '0';
if (a[i][j + 1] == '4')a[i][j + 1] = '1';
}else if(b[i][j]=='2') {
a[i][j] += 2;
if (a[i][j] == '3')a[i][j] = '0';
if (a[i][j] == '4')a[i][j] = '1';
a[i + 1][j + 1] += 2;
if (a[i + 1][j + 1] == '3')a[i + 1][j + 1] = '0';
if (a[i + 1][j + 1] == '4')a[i + 1][j + 1] = '1';
a[i + 1][j] += 1;
if (a[i + 1][j] == '3')a[i + 1][j] = '0';
if (a[i + 1][j] == '4')a[i + 1][j] = '1';
a[i][j + 1] += 1;
if (a[i][j + 1] == '3')a[i][j + 1] = '0';
if (a[i][j + 1] == '4')a[i][j + 1] = '1';
}
}
else if (a[i][j] == '1') {
if (b[i][j] == '2') {
a[i][j] += 1;
if (a[i][j] == '3')a[i][j] = '0';
if (a[i][j] == '4')a[i][j] = '1';
a[i + 1][j + 1] += 1;
if (a[i + 1][j + 1] == '3')a[i + 1][j + 1] = '0';
if (a[i + 1][j + 1] == '4')a[i + 1][j + 1] = '1';
a[i + 1][j] += 2;
if (a[i + 1][j] == '3')a[i + 1][j] = '0';
if (a[i + 1][j] == '4')a[i + 1][j] = '1';
a[i][j + 1] += 2;
if (a[i][j + 1] == '3')a[i][j + 1] = '0';
if (a[i][j + 1] == '4')a[i][j + 1] = '1';
}
else if (b[i][j] == '0') {
a[i][j] += 2;
if (a[i][j] == '3')a[i][j] = '0';
if (a[i][j] == '4')a[i][j] = '1';
a[i + 1][j + 1] += 2;
if (a[i + 1][j + 1] == '3')a[i + 1][j + 1] = '0';
if (a[i + 1][j + 1] == '4')a[i + 1][j + 1] = '1';
a[i + 1][j] += 1;
if (a[i + 1][j] == '3')a[i + 1][j] = '0';
if (a[i + 1][j] == '4')a[i + 1][j] = '1';
a[i][j + 1] += 1;
if (a[i][j + 1] == '3')a[i][j + 1] = '0';
if (a[i][j + 1] == '4')a[i][j + 1] = '1';
}
}
else {
if (b[i][j] == '0') {
a[i][j] += 1;
if (a[i][j] == '3')a[i][j] = '0';
if (a[i][j] == '4')a[i][j] = '1';
a[i + 1][j + 1] += 1;
if (a[i + 1][j + 1] == '3')a[i + 1][j + 1] = '0';
if (a[i + 1][j + 1] == '4')a[i + 1][j + 1] = '1';
a[i + 1][j] += 2;
if (a[i + 1][j] == '3')a[i + 1][j] = '0';
if (a[i + 1][j] == '4')a[i + 1][j] = '1';
a[i][j + 1] += 2;
if (a[i][j + 1] == '3')a[i][j + 1] = '0';
if (a[i][j + 1] == '4')a[i][j+1] = '1';
}
else if (b[i][j] == '1') {
a[i][j] += 2;
if (a[i][j] == '3')a[i][j] = '0';
if (a[i][j] == '4')a[i][j] = '1';
a[i + 1][j + 1] += 2;
if (a[i + 1][j + 1] == '3')a[i + 1][j + 1] = '0';
if (a[i + 1][j + 1] == '4')a[i + 1][j + 1] = '1';
a[i + 1][j] += 1;
if (a[i + 1][j] == '3')a[i + 1][j] = '0';
if (a[i + 1][j] == '4')a[i + 1][j] = '1';
a[i][j + 1] += 1;
if (a[i][j + 1] == '3')a[i][j + 1] = '0';
if (a[i][j + 1] == '4')a[i][j + 1] = '1';
}
}
}
}
}
for (int i = n; i > 1; i--) {
for (int j = m; j > 1; j--) {
if (a[i][j] != b[i][j]) {
if (a[i][j] == '0') {
if (b[i][j] == '1') {
a[i][j] += 1;
if (a[i][j] == '3')a[i][j] = '0';
if (a[i][j] == '4')a[i][j] = '1';
a[i - 1][j - 1] += 1;
if (a[i - 1][j - 1] == '3')a[i - 1][j - 1] = '0';
if (a[i - 1][j - 1] == '4')a[i - 1][j - 1] = '1';
a[i - 1][j] += 2;
if (a[i - 1][j] == '3')a[i - 1][j] = '0';
if (a[i - 1][j] == '4')a[i - 1][j] = '1';
a[i][j - 1] += 2;
if (a[i][j - 1] == '3')a[i][j - 1] = '0';
if (a[i][j - 1] == '4')a[i][j - 1] = '1';
}
else if (b[i][j] == '2') {
a[i][j] += 2;
if (a[i][j] == '3')a[i][j] = '0';
if (a[i][j] == '4')a[i][j] = '1';
a[i - 1][j - 1] += 2;
if (a[i - 1][j - 1] == '3')a[i - 1][j - 1] = '0';
if (a[i - 1][j - 1] == '4')a[i - 1][j -1] = '1';
a[i - 1][j] += 1;
if (a[i - 1][j] == '3')a[i - 1][j] = '0';
if (a[i - 1][j] == '4')a[i - 1][j] = '1';
a[i][j - 1] += 1;
if (a[i][j - 1] == '3')a[i][j - 1] = '0';
if (a[i][j - 1] == '4')a[i][j - 1] = '1';
}
}
else if (a[i][j] == '1') {
if (b[i][j] == '2') {
a[i][j] += 1;
if (a[i][j] == '3')a[i][j] = '0';
if (a[i][j] == '4')a[i][j] = '1';
a[i - 1][j - 1] += 1;
if (a[i - 1][j - 1] == '3')a[i - 1][j - 1] = '0';
if (a[i - 1][j - 1] == '4')a[i - 1][j - 1] = '1';
a[i - 1][j] += 2;
if (a[i - 1][j] == '3')a[i - 1][j] = '0';
if (a[i - 1][j] == '4')a[i - 1][j] = '1';
a[i][j - 1] += 2;
if (a[i][j - 1] == '3')a[i - 1][j] = '0';
if (a[i][j - 1] == '4')a[i - 1][j] = '1';
}
else if (b[i][j] == '0') {
a[i][j] += 2;
if (a[i][j] == '3')a[i][j] = '0';
if (a[i][j] == '4')a[i][j] = '1';
a[i - 1][j - 1] += 2;
if (a[i - 1][j - 1] == '3')a[i - 1][j - 1] = '0';
if (a[i - 1][j - 1] == '4')a[i - 1][j - 1] = '1';
a[i - 1][j] += 1;
if (a[i - 1][j] == '3')a[i - 1][j] = '0';
if (a[i - 1][j] == '4')a[i - 1][j] = '1';
a[i][j - 1] += 1;
if (a[i][j - 1] == '3')a[i][j - 1] = '0';
if (a[i][j - 1] == '4')a[i][j - 1] = '1';
}
}
else {
if (b[i][j] == '0') {
a[i][j] += 1;
if (a[i][j] == '3')a[i][j] = '0';
if (a[i][j] == '4')a[i][j] = '1';
a[i - 1][j - 1] += 1;
if (a[i - 1][j - 1] == '3')a[i - 1][j - 1] = '0';
if (a[i - 1][j - 1] == '4')a[i - 1][j - 1] = '1';
a[i - 1][j] += 2;
if (a[i - 1][j] == '3')a[i - 1][j] = '0';
if (a[i - 1][j] == '4')a[i - 1][j] = '1';
a[i][j - 1] += 2;
if (a[i][j - 1] == '3')a[i][j - 1] = '0';
if (a[i][j - 1] == '4')a[i][j-1] = '1';
}
else if (b[i][j] == '1') {
a[i][j] += 2;
if (a[i][j] == '3')a[i][j] = '0';
if (a[i][j] == '4')a[i][j] = '1';
a[i - 1][j - 1] += 2;
if (a[i - 1][j - 1] == '3')a[i - 1][j - 1] = '0';
if (a[i - 1][j - 1] == '4')a[i - 1][j-1] = '1';
a[i - 1][j] += 1;
if (a[i - 1][j] == '3')a[i - 1][j] = '0';
if (a[i - 1][j] == '4')a[i - 1][j] = '1';
a[i][j - 1] += 1;
if (a[i][j - 1] == '3')a[i][j - 1] = '0';
if (a[i][j - 1] == '4')a[i][j-1] = '1';
}
}
}
}
}
for (int i = 1; i <= n; i++) {
for (int j = 1; j <= m; j++) {
if (a[i][j] != b[i][j]) {
cout << "NO\n"; return;
}
}
}
cout << "YES\n";
}C. Have Your Cake and Eat It Too
题目大意
三个人吃蛋糕，每个人对蛋糕不同位置有自己的估价。
问如何分配蛋糕，使三个人获得的蛋糕价值均高于蛋糕价值和的三分之一。
思路
分类讨论：三个人拿蛋糕的顺序
-
abc
-
acb
-
bac
-
bca
-
cab
-
cba
代码实现
ll a[2100000];
ll b[2100000];
ll c[2100000];
void solve() {
ll n; cin >> n;
ll need = 0;
a[0] = b[0] = c[0] = 0;
for (int i = 1; i <= n; i++)cin >> a[i], need += a[i];
for (int i = 1; i <= n; i++)cin >> b[i];
for (int i = 1; i <= n; i++)cin >> c[i];
need = ceil(need / 3.0);
for (int i = 1; i <= n; i++)a[i] += a[i - 1];
for (int i = 1; i <= n; i++)b[i] += b[i - 1];
for (int i = 1; i <= n; i++)c[i] += c[i - 1];
a[n + 1] = b[n + 1] = c[n + 1] = 1e22;
//abc
ll id = lower_bound(a + 1, a + 1 + n, need) - a;
ll id2 = lower_bound(b + 1, b + 1 + n, b[id] + need) - b;
if (c[n] - c[id2] >= need) {
cout << 1 << " " << id << " " << id + 1 << " " << id2 << " " << id2 + 1 << " " << n << "\n";
return;
}
//acb
id = lower_bound(a + 1, a + 1 + n, need) - a;
id2 = lower_bound(c + 1, c + 1 + n, c[id] + need) - c;
if (b[n] - b[id2] >= need) {
cout << 1 << " " << id << " " << id2 + 1 << " " << n << " " << id + 1 << " " << id2 << "\n";
return;
}
//bca
id = lower_bound(b + 1, b + 1 + n, need) - b;
id2 = lower_bound(c + 1, c + 1 + n, c[id] + need) - c;
if (a[n] - a[id2] >= need) {
cout << id2 + 1 << " " << n << " " << 1 << " " << id << " " << id + 1 << " " << id2 << "\n";
return;
}
//bac
id = lower_bound(b + 1, b + 1 + n, need) - b;
id2 = lower_bound(a + 1, a + 1 + n, a[id] + need) - a;
if (c[n] - c[id2] >= need) {
cout << id + 1 << " " << id2 << " " << 1 << " " << id << " " << id2 + 1 << " " << n << "\n";
return;
}
//cab
id = lower_bound(c + 1, c + 1 + n, need) - c;
id2 = lower_bound(a + 1, a + 1 + n, a[id] + need) - a;
if (b[n] - b[id2] >= need) {
cout << id + 1 << " " << id2 << " " << id2 + 1 << " " << n <<" " << 1 << " " << id << "\n";
return;
}
//cba
id = lower_bound(c + 1, c + 1 + n, need) - c;
id2 = lower_bound(b + 1, b + 1 + n, b[id] + need) - b;
if (a[n] - a[id2] >= need) {
cout << id2 + 1 << " " << n << " " << id + 1 << " " << id2 << " " << 1 << " " << id << "\n";
return;
}
cout << "-1\n";
}