A、AOE还是单体？
发现的时候直接单体伤害最优。那么其他时候就是调整
让它和
相同即可，剩下的单体即可。
#include <bits/stdc++.h>
#pragma GCC optimize(2)
#pragma GCC optimize(3)
using namespace std;
#define js ios::sync_with_stdio(false);cin.tie(0); cout.tie(0)
typedef long long ll;
const ll MOD = 1e9 + 7;
inline ll read() { ll s = 0, w = 1; char ch = getchar(); while (ch < 48 || ch > 57) { if (ch == '-') w = -1; ch = getchar(); } while (ch >= 48 && ch <= 57) s = (s << 1) + (s << 3) + (ch ^ 48), ch = getchar(); return s * w; }
inline void write(ll x) { if (!x) { putchar('0'); return; } char F[200]; ll tmp = x > 0 ? x : -x; if (x < 0)putchar('-'); int cnt = 0; while (tmp > 0) { F[cnt++] = tmp % 10 + '0'; tmp /= 10; } while (cnt > 0)putchar(F[--cnt]); }
inline ll gcd(ll x, ll y) { return y ? gcd(y, x % y) : x; }
ll qpow(ll a, ll b) { ll ans = 1; while (b) { if (b & 1) ans *= a; b >>= 1; a *= a; } return ans; } ll qpow(ll a, ll b, ll mod) { ll ans = 1; while (b) { if (b & 1)(ans *= a) %= mod; b >>= 1; (a *= a) %= mod; }return ans % mod; }
inline int lowbit(int x) { return x & (-x); }
const double eps = 1e-9;
ll a[200000 + 7];
int main() {
ll n = read(), x = read(), sum = 0;
for (int i = 0; i < n; ++i) {
a[i] = read();
sum += a[i];
}
if (x >= n) {
write(sum), putchar(32);
}
else {
sort(a, a + n);
ll ans = 0;
ans += (a[n - x - 1] * x);
for (int i = n - x; i < n; ++i) ans += (a[i] - a[n - x - 1]);
write(ans), putchar(32);
}
return 0;
} B、k-size字符串
待补
C、白魔法师
待补
D、抽卡
首先看过不掉样例的想法，认为答案是累乘的。其实很简单，你算的是抽到n张卡，我们只是最少1张卡即可。
正解：累乘再用
减掉，算到拿不到卡概率，减掉即可。
#include <bits/stdc++.h>
#pragma GCC optimize(2)
#pragma GCC optimize(3)
using namespace std;
#define js ios::sync_with_stdio(false);cin.tie(0); cout.tie(0)
typedef long long ll;
const ll MOD = 1e9 + 7;
inline ll read() { ll s = 0, w = 1; char ch = getchar(); while (ch < 48 || ch > 57) { if (ch == '-') w = -1; ch = getchar(); } while (ch >= 48 && ch <= 57) s = (s << 1) + (s << 3) + (ch ^ 48), ch = getchar(); return s * w; }
inline void write(ll x) { if (!x) { putchar('0'); return; } char F[200]; ll tmp = x > 0 ? x : -x; if (x < 0)putchar('-'); int cnt = 0; while (tmp > 0) { F[cnt++] = tmp % 10 + '0'; tmp /= 10; } while (cnt > 0)putchar(F[--cnt]); }
inline ll gcd(ll x, ll y) { return y ? gcd(y, x % y) : x; }
ll qpow(ll a, ll b) { ll ans = 1; while (b) { if (b & 1) ans *= a; b >>= 1; a *= a; } return ans; } ll qpow(ll a, ll b, ll mod) { ll ans = 1; while (b) { if (b & 1)(ans *= a) %= mod; b >>= 1; (a *= a) %= mod; }return ans % mod; }
inline int lowbit(int x) { return x & (-x); }
const int N = 1e5 + 7;
ll a[N], b[N];
int main() {
int n = read();
for (int i = 1; i <= n; ++i) a[i] = read();
for (int i = 1; i <= n; ++i) b[i] = read();
ll up = 1, down = 1;
for (int i = 1; i <= n; ++i)
(up *= a[i] - b[i]) %= MOD, (down *= a[i]) %= MOD;
up = (down - up + MOD) % MOD;
write(up * qpow(down, MOD - 2, MOD) % MOD), putchar(32);
return 0;
} E、点击消除
手写栈，的栈不好操作，以为后序输出需要倒叙。遇到和栈顶相同，出栈即可，否则进栈。
#include <bits/stdc++.h>
#pragma GCC optimize(2)
#pragma GCC optimize(3)
using namespace std;
#define js ios::sync_with_stdio(false);cin.tie(0); cout.tie(0)
typedef long long ll;
const ll MOD = 1e9 + 7;
inline ll read() { ll s = 0, w = 1; char ch = getchar(); while (ch < 48 || ch > 57) { if (ch == '-') w = -1; ch = getchar(); } while (ch >= 48 && ch <= 57) s = (s << 1) + (s << 3) + (ch ^ 48), ch = getchar(); return s * w; }
inline void write(ll x) { if (!x) { putchar('0'); return; } char F[200]; ll tmp = x > 0 ? x : -x; if (x < 0)putchar('-'); int cnt = 0; while (tmp > 0) { F[cnt++] = tmp % 10 + '0'; tmp /= 10; } while (cnt > 0)putchar(F[--cnt]); }
inline ll gcd(ll x, ll y) { return y ? gcd(y, x % y) : x; }
ll qpow(ll a, ll b) { ll ans = 1; while (b) { if (b & 1) ans *= a; b >>= 1; a *= a; } return ans; } ll qpow(ll a, ll b, ll mod) { ll ans = 1; while (b) { if (b & 1)(ans *= a) %= mod; b >>= 1; (a *= a) %= mod; }return ans % mod; }
inline int lowbit(int x) { return x & (-x); }
int main() {
js;
char a[300000 + 7], ch;
int top = 0;
while (cin >> ch) {
if (!top) a[top++] = ch;
else {
if (ch == a[top - 1]) --top;
else a[top++] = ch;
}
}
if (!top) {
puts("0");
return 0;
}
for (int i = 0; i < top; ++i)
cout << a[i];
return 0;
} F、疯狂的自我检索者
签到题，最小全1，最大全5即可
#include <bits/stdc++.h>
#pragma GCC optimize(2)
#pragma GCC optimize(3)
using namespace std;
#define js ios::sync_with_stdio(false);cin.tie(0); cout.tie(0)
typedef long long ll;
const ll MOD = 1e9 + 7;
inline ll read() { ll s = 0, w = 1; char ch = getchar(); while (ch < 48 || ch > 57) { if (ch == '-') w = -1; ch = getchar(); } while (ch >= 48 && ch <= 57) s = (s << 1) + (s << 3) + (ch ^ 48), ch = getchar(); return s * w; }
inline void write(ll x) { if (!x) { putchar('0'); return; } char F[200]; ll tmp = x > 0 ? x : -x; if (x < 0)putchar('-'); int cnt = 0; while (tmp > 0) { F[cnt++] = tmp % 10 + '0'; tmp /= 10; } while (cnt > 0)putchar(F[--cnt]); }
inline ll gcd(ll x, ll y) { return y ? gcd(y, x % y) : x; }
ll qpow(ll a, ll b) { ll ans = 1; while (b) { if (b & 1) ans *= a; b >>= 1; a *= a; } return ans; } ll qpow(ll a, ll b, ll mod) { ll ans = 1; while (b) { if (b & 1)(ans *= a) %= mod; b >>= 1; (a *= a) %= mod; }return ans % mod; }
inline int lowbit(int x) { return x & (-x); }
const int N = 2e5 + 7;
int a[N];
int main() {
int n = read(), m = read();
int sum = 0;
for (int i = 1; i <= n - m; ++i) sum += read();
printf("%.5f %.5f\n", (sum + m) * 1.0 / n, (sum + m * 5) * 1.0 / n);
return 0;
} G、解方程
出题人放过了我们的二分，却让我们死在了精度……这也学到个知识，二分次数到了一定时候直接出循环即可解决区间不动的问题。
）不然开局30分钟就A了的小声咕咕
#include <bits/stdc++.h>
#pragma GCC optimize(2)
#pragma GCC optimize(3)
using namespace std;
#define js ios::sync_with_stdio(false);cin.tie(0); cout.tie(0)
typedef long long ll;
const ll MOD = 1e9 + 7;
inline ll read() { ll s = 0, w = 1; char ch = getchar(); while (ch < 48 || ch > 57) { if (ch == '-') w = -1; ch = getchar(); } while (ch >= 48 && ch <= 57) s = (s << 1) + (s << 3) + (ch ^ 48), ch = getchar(); return s * w; }
inline void write(ll x) { if (!x) { putchar('0'); return; } char F[200]; ll tmp = x > 0 ? x : -x; if (x < 0)putchar('-'); int cnt = 0; while (tmp > 0) { F[cnt++] = tmp % 10 + '0'; tmp /= 10; } while (cnt > 0)putchar(F[--cnt]); }
inline ll gcd(ll x, ll y) { return y ? gcd(y, x % y) : x; }
ll qpow(ll a, ll b) { ll ans = 1; while (b) { if (b & 1) ans *= a; b >>= 1; a *= a; } return ans; } ll qpow(ll a, ll b, ll mod) { ll ans = 1; while (b) { if (b & 1)(ans *= a) %= mod; b >>= 1; (a *= a) %= mod; }return ans % mod; }
inline int lowbit(int x) { return x & (-x); }
const double eps = 1e-7;
double a, b, c;
bool calc(double x) {
return pow(x, a) + b * log(x) > c;
}
int main() {
scanf("%lf %lf %lf", &a, &b, &c);
double l = eps, r = 1e9;
int cnt=0;
while (fabs(r - l) > eps) {
++cnt;
double mid = (r + l) / 2.0;
if (calc(mid)) r = mid;
else l = mid;
if(cnt>1000) break;
}
printf("%.14f\n", l);
return 0;
} H、神奇的字母（二）
一血，哎嘿舒服了。直接通过输入到文件结束一直累加字母，再找到最大的那个即可。
#include <bits/stdc++.h>
#pragma GCC optimize(2)
#pragma GCC optimize(3)
using namespace std;
#define js ios::sync_with_stdio(false);cin.tie(0); cout.tie(0)
typedef long long ll;
const ll MOD = 1e9 + 7;
inline ll read() { ll s = 0, w = 1; char ch = getchar(); while (ch < 48 || ch > 57) { if (ch == '-') w = -1; ch = getchar(); } while (ch >= 48 && ch <= 57) s = (s << 1) + (s << 3) + (ch ^ 48), ch = getchar(); return s * w; }
inline void write(ll x) { if (!x) { putchar('0'); return; } char F[200]; ll tmp = x > 0 ? x : -x; if (x < 0)putchar('-'); int cnt = 0; while (tmp > 0) { F[cnt++] = tmp % 10 + '0'; tmp /= 10; } while (cnt > 0)putchar(F[--cnt]); }
inline ll gcd(ll x, ll y) { return y ? gcd(y, x % y) : x; }
ll qpow(ll a, ll b) { ll ans = 1; while (b) { if (b & 1) ans *= a; b >>= 1; a *= a; } return ans; } ll qpow(ll a, ll b, ll mod) { ll ans = 1; while (b) { if (b & 1)(ans *= a) %= mod; b >>= 1; (a *= a) %= mod; }return ans % mod; }
inline int lowbit(int x) { return x & (-x); }
int a[30];
int main() {
js;
char ch;
while (cin >> ch) {
if (ch >= 'a' && ch <= 'z') ++a[ch - 'a'];
}
//返回地址 - a 就是偏移量
cout << char('a' + max_element(a, a + 30) - a) << endl;
return 0;
} I、十字爆破
开两个数组，记录行和以及列和，在开个动态数组记录每个位置的元素。
最后用行和加列和减当前位置元素输出即可。
#include <bits/stdc++.h>
#pragma GCC optimize(2)
#pragma GCC optimize(3)
using namespace std;
#define js ios::sync_with_stdio(false);cin.tie(0); cout.tie(0)
typedef long long ll;
const ll MOD = 1e9 + 7;
inline ll read() { ll s = 0, w = 1; char ch = getchar(); while (ch < 48 || ch > 57) { if (ch == '-') w = -1; ch = getchar(); } while (ch >= 48 && ch <= 57) s = (s << 1) + (s << 3) + (ch ^ 48), ch = getchar(); return s * w; }
inline void write(ll x) { if (!x) { putchar('0'); return; } char F[200]; ll tmp = x > 0 ? x : -x; if (x < 0)putchar('-'); int cnt = 0; while (tmp > 0) { F[cnt++] = tmp % 10 + '0'; tmp /= 10; } while (cnt > 0)putchar(F[--cnt]); }
inline ll gcd(ll x, ll y) { return y ? gcd(y, x % y) : x; }
ll qpow(ll a, ll b) { ll ans = 1; while (b) { if (b & 1) ans *= a; b >>= 1; a *= a; } return ans; } ll qpow(ll a, ll b, ll mod) { ll ans = 1; while (b) { if (b & 1)(ans *= a) %= mod; b >>= 1; (a *= a) %= mod; }return ans % mod; }
inline int lowbit(int x) { return x & (-x); }
const int N = 1e6 + 7;
ll r[N], c[N]; //行 列
vector<int> p;
int main() {
int n = read(), m = read();
for (int i = 1; i <= n; ++i) {
for (int j = 1; j <= m; ++j) {
int tmp = read();
p.push_back(tmp);
r[i] += tmp;
c[j] += tmp;
}
}
for (int i = 1; i <= n; ++i)
for (int j = 1; j <= m; ++j)
printf("%lld%c", r[i] + c[j] - p[(i - 1) * m + j - 1], " \n"[j == m]);
return 0;
}
J、异或和之和
看题目发现，只有三个数异或出来是1，才有价值，那么根据异或的性质可以发现 或者
那么题目就变成了从这一位选三个一的方案数加这一位选一个一再选两个零的方案数之和。每一位的一个数预处理即可，零个数N减掉一个数即可
#include <bits/stdc++.h>
#pragma GCC optimize(2)
#pragma GCC optimize(3)
using namespace std;
#define js ios::sync_with_stdio(false);cin.tie(0); cout.tie(0)
typedef long long ll;
const ll MOD = 1e9 + 7;
inline ll read() { ll s = 0, w = 1; char ch = getchar(); while (ch < 48 || ch > 57) { if (ch == '-') w = -1; ch = getchar(); } while (ch >= 48 && ch <= 57) s = (s << 1) + (s << 3) + (ch ^ 48), ch = getchar(); return s * w; }
inline void write(ll x) { if (!x) { putchar('0'); return; } char F[200]; ll tmp = x > 0 ? x : -x; if (x < 0)putchar('-'); int cnt = 0; while (tmp > 0) { F[cnt++] = tmp % 10 + '0'; tmp /= 10; } while (cnt > 0)putchar(F[--cnt]); }
inline ll gcd(ll x, ll y) { return y ? gcd(y, x % y) : x; }
ll qpow(ll a, ll b) { ll ans = 1; while (b) { if (b & 1) ans *= a; b >>= 1; a *= a; } return ans; } ll qpow(ll a, ll b, ll mod) { ll ans = 1; while (b) { if (b & 1)(ans *= a) %= mod; b >>= 1; (a *= a) %= mod; }return ans % mod; }
inline int lowbit(int x) { return x & (-x); }
ll C2(ll x) { return x * (x - 1) / 2 % MOD; } //组合数没有小数所以不求逆元也没关系
ll C3(ll x) { return x * (x - 1) * (x - 2) / 6 % MOD; }
ll cnt[70];
ll main() {
int n = read();
for (int i = 1; i <= n; ++i) {
ll tmp = read(), b = 0;
while (tmp) cnt[b++] += tmp & 1, tmp >>= 1;
}
ll base = 1, ans = 0;
for (int i = 0; i < 64; ++i) {
(ans += (C3(cnt[i]) + cnt[i] * C2(n - cnt[i]) % MOD) % MOD * base % MOD) %= MOD;
base = (base << 1) % MOD;
}
write(ans), putchar(32);
return 0;
} 
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