在家都变的懒惰了，好久没写题解了，补补CF
模拟 A - Wet Shark and Odd and Even
#include <bits/stdc++.h> typedef long long ll;
const int N = 1e5 + 5;
const int INF = 0x3f3f3f3f; int main(void) {
std::vector<int> vec;
int n; scanf ("%d", &n);
ll sum = 0;
for (int x, i=1; i<=n; ++i) {
scanf ("%d", &x);
sum += x;
if (x & 1) vec.push_back (x);
}
std::sort (vec.begin (), vec.end ());
int sz = vec.size ();
if (sz > 0 && (sz & 1)) sum -= vec[0];
printf ("%I64d\n", sum); return 0;
}
开始想错了，当成斜率相等的。还好1000范围不大，统计矩阵每条对角线上的个数加点小优化就过了，代码丑。。。
#include <bits/stdc++.h> typedef long long ll;
const int N = 2e5 + 5;
bool vis[2][1005][1005];
int b[1005][1005];
std::pair<int, int> a[N]; ll cal(int x) {
return 1ll * x * (x - 1) / 2;
}
int get_num1(int x, int y) {
int xx = x, yy = y;
int ret = 0;
while (xx >= 1 && yy >= 1) {
if (b[xx][yy]) ret++, vis[1][xx][yy] = true;
xx--; yy--;
}
xx = x + 1, yy = y + 1;
while (xx <= 1000 && yy <= 1000) {
if (b[xx][yy]) ret++, vis[1][xx][yy] = true;
xx++; yy++;
}
return ret;
} int get_num0(int x, int y) {
int xx = x, yy = y;
int ret = 0;
while (xx <= 1000 && yy >= 1) {
if (b[xx][yy]) ret++, vis[0][xx][yy] = true;
xx++; yy--;
}
xx = x - 1, yy = y + 1;
while (xx >= 1 && yy <= 1000) {
if (b[xx][yy]) ret++, vis[0][xx][yy] = true;
xx--; yy++;
}
return ret;
} int main(void) {
int n; scanf ("%d", &n);
for (int i=0; i<n; ++i) {
scanf ("%d%d", &a[i].first, &a[i].second);
b[a[i].first][a[i].second] = 1;
}
ll ans = 0;
for (int i=0; i<n; ++i) {
int x = a[i].first, y = a[i].second;
if (!vis[0][x][y]) {
ans += cal (get_num0 (x, y));
vis[0][x][y] = true;
}
if (!vis[1][x][y]) {
ans += cal (get_num1 (x, y));
vis[1][x][y] = true;
}
}
printf ("%I64d\n", ans); return 0;
}
E = sum (1000 * 乘积是p的倍数的概率)
#include <bits/stdc++.h> typedef long long ll;
const int N = 1e5 + 5;
const int EPS = 1e-8;
int l[N], r[N];
double pos[N]; int main(void) {
int n, p; scanf ("%d%d", &n, &p);
for (int i=1; i<=n; ++i) {
scanf ("%d%d", &l[i], &r[i]);
pos[i] = 1.0 * (r[i] / p - ((l[i]-1) / p)) / (r[i] - l[i] + 1);
}
l[0] = l[n], r[0] = r[n], pos[0] = pos[n];
double ans = 0;
for (int i=0; i<=n; ++i) {
if (i < n) ans += pos[i] * 1.0 + (1.0 - pos[i]) * pos[i+1];
if (i > 0) ans += pos[i] * 1.0 + (1.0 - pos[i]) * pos[i-1];
}
ans *= 1000;
printf ("%.8f\n", ans); return 0;
}
数学(浮点) D - Rat Kwesh and Cheese
都取log，用long double，精度逆天！powl (): pow的long double版
C语言里对float类型数据的表示范围为-3.4*10^38～+3.4*10^38。double为-1.7*10^-308~1.7*10^308，long double为-1.2*10^-4932~1.2*10^4932.
|
类型 |
比特（位）数 |
有效数字 |
数值范围 |
|
float |
32 |
6~7 |
-3.4*10^38～+3.4*10^38 |
|
double |
64 |
15~16 |
-1.7*10^-308~1.7*10^308 |
|
long double |
128/ |
18~19 |
-1.2*10^-4932~1.2*10^4932 |
究竟如何计算该范围，分析如下：
对于单精度浮点数（float）来说，符号位一位，指数位8位，尾数23位。指数能够表示的指数范围为-128~127。尾数为23位。
#include <bits/stdc++.h>
std::string ans[12] = {
"x^y^z", "x^z^y", "(x^y)^z", "(x^z)^y", "y^x^z", "y^z^x",
"(y^x)^z", "(y^z)^x", "z^x^y", "z^y^x", "(z^x)^y", "(z^y)^x"
};
typedef long double ldouble;
const double EPS = 1e-10;
ldouble best;
int id;
bool better(ldouble val) {
if (fabs (val - best) < EPS) return false;
else if (best < val) {
best = val;
return true;
}
return false;
}
void try2(ldouble x, ldouble y, ldouble z, int pos) {
//(x ^ y) ^ z
ldouble val = z * y * log (x);
if (better (val)) id = pos;
}
void try1(ldouble x, ldouble y, ldouble z, int pos) {
//x ^ y ^ z
ldouble val = powl (y, z) * log (x);
if (better (val)) id = pos;
}
int main(void) {
ldouble x, y, z; std::cin >> x >> y >> z;
best = -1e12; id = -1;
try1 (x, y, z, 0);
try1 (x, z, y, 1);
try2 (x, y, z, 2);
try2 (x, z, y, 3);
try1 (y, x, z, 4);
try1 (y, z, x, 5);
try2 (y, x, z, 6);
try2 (y, z, x, 7);
try1 (z, x, y, 8);
try1 (z, y, x, 9);
try2 (z, x, y, 10);
try2 (z, y, x, 11);
std::cout << ans[id] << '\n';
return 0;
}
数位DP(矩阵快速幂优化) E - Wet Shark and Blocks
题意：b组相同的n个数字，每组选择一个数字合并起来%x == k的方案数。
分析：dp思路，dp[i][(j*10+t)%x] += dp[i-1][j]。但是b太大了选择用矩阵快速幂优化，构造矩阵m[i][j]表示从i到j的方案数，答案就是m[0][k]，那么base矩阵就是小于x的某个数字转移到根据给出n个数字转移到另一个数字y的方案数。
UPD：
寒假排位赛出了这题，又不会做。
构造x*x大小的矩阵，m[i][j]表示从i(%x)到j(%x)的方案数，状态转移用矩阵乘法：i->j1,j2,j3...->k => i->k
#include <bits/stdc++.h>
using namespace std; typedef long long ll;
typedef vector<ll> Vec;
typedef vector<Vec> Mat;
const int N = 5e4 + 5;
const int MOD = 1e9 + 7;
int cnt[10];
int n, b, k, x; void add_mod(ll &a, ll b) {
a += b;
if (a >= MOD) a -= MOD;
} Mat matrix_mul(const Mat &A, const Mat &B) {
Mat ret(A.size(), Vec(B[0].size()));
for (int i=0; i<A.size(); ++i)
for (int j=0; j<A[0].size(); ++j) if (A[i][j])
for (int k=0; k<B[0].size(); ++k) if (B[j][k])
add_mod(ret[i][k], A[i][j]*B[j][k]%MOD);
return ret;
} Mat matrix_pow(Mat X, int n) {
Mat ret(X.size(), Vec(X.size()));
for (int i=0; i<X.size(); ++i) ret[i][i] = 1;
for (; n; n>>=1) {
if (n & 1) ret = matrix_mul(ret, X);
X = matrix_mul(X, X);
}
return ret;
} int main() {
scanf("%d%d%d%d", &n, &b, &k, &x);
for (int i=1; i<=n; ++i) {
int d;
scanf("%d", &d);
cnt[d]++;
}
Mat ans(100, Vec(100));
for (int i=0; i<x; ++i) {
for (int j=1; j<=9; ++j) {
ans[i][(i*10+j)%x] += cnt[j];
}
}
ans = matrix_pow(ans, b);
printf("%I64d\n", ans[0][k]);
return 0;
}