Codeforces Round #341 (Div. 2)

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;
}

暴力 B - Wet Shark and Bishops

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;
}

期望 C - Wet Shark and Flowers

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;
}