A-Bingbong的化学世界
#include<bits/stdc++.h>
using namespace std;
const int maxn = 1001;
int a[maxn];
int main() {
string t = "...|...";
vector<string> x(6);
for (auto &i: x) cin >> i;
if (x.front() == t) {
if (x.back() == t) cout << "p";
else cout << "m";
} else cout << "o";
return 0;
}
B-Bingbong的数数世界
#include<bits/stdc++.h>
using namespace std;
void solve(){
int n;
cin >> n;
if( n % 4 == 0 ) cout << "Bong\n";
else cout << "Bing\n";
}
int main() {
int TC;
for (cin >> TC; TC; TC--)
solve();
return 0;
}
C-Bingbong的蛋仔世界
#include<bits/stdc++.h>
using namespace std;
using pii = pair<int, int>;
int main() {
ios::sync_with_stdio(false), cin.tie(nullptr);
int n, m, k;
cin >> n >> m >> k;
int lx = 1, rx = n, ly = 1, ry = m, mx = n / 2 + 1, my = m / 2 + 1;
vector<pii> a(k);
for (auto &[x, y]: a) cin >> x >> y;
int res = 0;
for (int dx, dy; not a.empty();) {
if (lx != mx) lx++, rx--;
if (ly != my) ly++, ry--;
vector<pii> b;
for (const auto &[x, y]: a) {
if (x == mx and y == my) {
res++;
continue;
}
if (x != mx and ly <= y and y <= ry) {
if (x < mx) dx = x + 1;
else dx = x - 1;
if (lx <= dx and dx <= rx) b.emplace_back(dx, y);
} else if (y != my and lx <= x and x <= rx) {
if (y < my) dy = y + 1;
else dy = y - 1;
if (ly <= dy and dy <= ry) b.emplace_back(x, dy);
}
}
a.swap(b);
}
cout << res << "\n";
return 0;
}
D-Bingbong的奇偶世界
#include <bits/stdc++.h>
using namespace std;
using i32 = int32_t;
using i64 = long long;
#define int i64
using vi = vector<int>;
const int mod = 1e9 + 7;
int power(int x, int y) {
int ans = 1;
while (y) {
if (y & 1) ans = ans * x % mod;
x = x * x % mod, y /= 2;
}
return ans;
}
i32 main() {
string s;
int n;
cin >> n >> s;
int res = 0;
for (int i = 0, x, cnt = 0; i < n; i++) {
x = s[i] - '0';
if (x % 2 == 0) res = (res + cnt + 1) % mod;
cnt = cnt * 2 % mod;
if (x != 0) cnt = (cnt + 1) % mod;
}
cout << res << "\n";
return 0;
}
F-Bingbong的幻想世界
做法就是进行拆位,拆位后用前后缀和统计一下每个1 可以产生多少次贡献。
#include<bits/stdc++.h>
using namespace std;
using i32 = int32_t;
using i64 = long long;
#define int long long
using vi = vector<int>;
const int mod = 1e9 + 7;
i32 main() {
ios::sync_with_stdio(false), cin.tie(nullptr);
int n;
cin >> n;
vector a(20, vi(n + 1));
for (int i = 1, x; i <= n; i++) {
cin >> x;
for (int j = 0; j < 20; j++) {
if (x & 1) a[j][i] = 1;
x >>= 1;
}
}
int res = 0;
for (int l = 0, pre, suf; l < 20; l++) {
pre = suf = 0;
int ans = 0;
for (int i = 1; i <= n; i++)
if (a[l][i] == 0) suf += (n - i + 1);
for (int i = 1; i <= n; i++) {
if (a[l][i] == 0)
pre += i, suf -= (n - i + 1);
else
ans = (ans + pre * (n - i + 1) % mod + i * suf % mod) % mod;
}
res = (res + (1ll << l) * ans) % mod;
}
cout << res * 2 % mod << "\n";
return 0;
}
标签:cnt,int,res,cin,牛客,小白月赛,91,using,mod
From: https://www.cnblogs.com/PHarr/p/18175571