Educational Codeforces Round 118 (Rated for Div. 2)
好久没更新了。。
A
#include
using namespace std;
typedef long long ll;
#define P pair
#define x first
#define y second
const int N = 5e5 + 10;
string a, b;
int main() {
int T;
cin >> T;
while (T--) {
int l1, l2;
cin >> a >> l1 >> b >> l2;
if (a.length() + l1 > b.length() + l2) {
puts(">");
} else if (a.length() + l1 < b.length() + l2) {
puts("<");
} else {
// for(int i=0;i b[p1]) {
puts(">");
break;
} else if (a[p1] < b[p1]) {
puts("<");
break;
}
p1++, p2++;
}
// cout << p1 << " " << p2 << endl;
if (p1 < a.length() && p2 == b.length()) {
bool f = 0;
while (p1 < a.length()) {
if (a[p1++] != '0') {
puts(">");
f = 1;
break;
}
}
if (!f)
puts("=");
} else if (p1 == a.length() && p2 < b.length()) {
bool f = 0;
while (p2 < b.length()) {
if (b[p2++] != '0') {
puts("<");
f = 1;
break;
}
}
if (!f)
puts("=");
} else if (p1 == a.length() && p2 == b.length())
puts("=");
}
}
}
B
不难发现可以用最小的数去模
麻了，一直RE，感觉自己是个傻子
#include
using namespace std;
typedef long long ll;
#define P pair
#define x first
#define y second
const int N = 1e6 + 10;
int a[N], n;
bool pos[N << 1], vis[N << 1];
int main() {
int T;
cin >> T;
while (T--) {
cin >> n;
int mn = 0;
for (int i = 0; i < n; i++) {
cin >> a[i];
mn = max(mn, a[i]);
pos[a[i]] = 1;
}
sort(a, a + n);
int need = n / 2;
for (int k = 1; k < n && need--; k++) {
// for (int i = 0; i +k < n && need; i++) {
// // cout<<"cjeck: "< C
#include
using namespace std;
typedef long long ll;
const int N = 105;
int a[N], n;
ll need;
bool judge(ll k) {
ll len = 0;
int pre = 0;
for (int i = 0; i < n - 1; i++) {
len += min(k,1ll*a[i+1] - a[i]);
// pre = a[i];
}
len += k;
// cout<= need)
return true;
return false;
}
int main() {
int T;
cin >> T;
while (T--) {
cin >> n >> need;
for(int i=0;i>a[i];
// judge(4);
ll l = 0, r = 1e18 + 1, mid, ans;
while (l <= r) {
mid = (l + r) >> 1;
if (judge(mid)) {
r = mid - 1;
ans = mid;
} else {
l = mid + 1;
}
}
cout< D
计数。。还不错的题
#include
using namespace std;
typedef long long ll;
const int N = 5e5 + 10;
const int mod = 998244353;
ll dp[N], dp2[N], a[N];
int main() {
int n, T;
cin >> T;
while (T--) {
cin >> n;
for (int i = 0; i <= n; i++) dp[i] = dp2[i] = 0;
dp[0] = 1;
int x;
for (int i = 0; i < n; i++) {
cin >> x;
dp[x + 1] = (dp[x + 1]+ dp[x + 1]+dp[x])%mod;
if (x > 0) dp2[x - 1] = (dp2[x - 1]+ dp2[x - 1]+dp[x-1])%mod;
dp2[x + 1] = (dp2[x + 1] + dp2[x + 1]) % mod;
}
int ans = 0;
for (int i = 0; i <= n; i++) ans = (ans + dp[i]) % mod;
for (int i = 0; i <= n; i++) ans = (ans + dp2[i]) % mod;
cout << (ans + mod -1 )%mod<< endl;
}
}
E
bfs水题，这咋放到E的。。
#include
#define P pair
#define x first
#define y second
#define endl '\n'
using namespace std;
const int N = 1e6;
vector> mp, vis;
int n, m, sx, sy;
int dx[] = {0, 0, 1, -1};
int dy[] = {1, -1, 0, 0};
bool check(int x, int y) {
return (1 <= x && x <= n && 1 <= y && y <= m && mp[x][y] && !vis[x][y]);
}
int main() {
ios::sync_with_stdio(0);
cin.tie(0);
cout.tie(0);
int T;
cin >> T;
while (T--) {
cin >> n >> m;
mp.resize(n + 1);
vis.resize(n + 1);
char tmp;
for (int i = 1; i <= n; i++) {
mp[i].resize(m + 1);
vis[i].resize(m + 1);
for (int j = 1; j <= m; j++) {
cin >> tmp;
if (tmp == 'L') {
sx = i, sy = j;
}
mp[i][j] = tmp == '.' || tmp == 'L' ? 1 : 0;
vis[i][j] = 0;
}
}
queue q;
q.push(P(sx, sy));
vis[sx][sy] = 1;
while (q.size()) {
P now = q.front();
q.pop();
vector
ans;
for (int i = 0; i < 4; i++) {
int nx = now.x + dx[i], ny = now.y + dy[i];
if (check(nx, ny)) {
ans.push_back(P(nx, ny));
}
}
// cout << "walk: " << now.x << " " << now.y << " " << ans.size()
// << endl;
if (ans.size() <= 1 || (now.x == sx && now.y == sy)) {
vis[now.x][now.y] = true;
for (auto p : ans) {
if (!vis[p.x][p.y]) q.push(P(p.x, p.y));
}
}
}
for (int i = 1; i <= n; i++) {
for (int j = 1; j <= m; j++) {
if (i == sx && j == sy) {
cout << 'L';
continue;
}
cout << (vis[i][j] ? '+' : (mp[i][j] ? '.' : '#'));
}
cout << endl;
}
}
}