动态规划部分题目代码记录

A

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#include<iostream>
#include<algorithm>
using namespace std;
const int N = 105;
#define ll long long
ll t, shu[N], n;

int main() {
    cin >> t;
    shu[1] = 1; shu[0] = 1;
    for (int i = 2; i < 82; i++)
        shu[i] = shu[i - 1] + shu[i - 2];
    //cout << shu[80] << endl;
    while (t--) {
        cin >> n;
        cout << shu[n] << endl;
    }
    return 0;
}

B

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#include<iostream>
#include<algorithm>
using namespace std;
const int N = 1e5 + 5, mod = 1e9 + 7;
int n, t;
long long shu[N];

void  chu() {
    shu[1] = 3; shu[2] = 9; shu[3] = 24;
    for (int i = 4; i < N; i++) {
        shu[i] = shu[i - 1] * 3 % mod;
        shu[i] -= 2 * shu[i - 3];
        shu[i] = (shu[i] +2* mod) % mod;
    }
}

int main() {
    chu();
    cin >> t;
    while (t--) {
        cin >> n;
        cout << shu[n] << endl;
    }
    return 0;
}

C

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#include<iostream>
#include<algorithm>
using namespace std;
#define ll long long
const int n = 1e5 + 5, mod = 1e9 + 7;
ll n, t, dp[n];

void  chu() {
    for (ll i = 3; i < n; i++) {
        if (i & 1)
            dp[i] = dp[i / 2] + dp[i / 2 + 1] + 1;
        else
            dp[i] = dp[i / 2] * 2;
    }
}

ll findd(ll n) {
    if (n <= 100000)  return dp[n];
    if (n & 1)
        return findd(n / 2) + findd(n / 2 + 1) + 1;
    else
        return 2 * findd(n / 2);
}

int main() {
    chu();
    cin >> t;
    if (t == 1) {
        cin >> n;
        cout << findd(n) << endl;
    }
    else {
        while (t--) {
            cin >> n;
            cout << dp[n] << endl;
        }
    }
    return 0;
}

D

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#include<iostream>
#include<algorithm>
using namespace std;
const int N = 1005;
int n, dp[N][N], a[N][N];

int main()
{
    cin >> n;
    for (int i = 1; i <= n; i++)
        for (int j = 1; j <= i; j++)
            cin >> a[i][j];
    for (int i = 1; i <= n; i++)
        for (int j = 0; j <= i + 1; j++)
            dp[i][j] = -1e9;
    dp[1][1] = a[1][1];
    for (int i = 2; i <= n; i++)
        for (int j = 1; j <= i; j++)
            dp[i][j] = max(dp[i - 1][j - 1] + a[i][j], dp[i - 1][j] + a[i][j]);
    int rex = -1e9;
    for (int i = 1; i <= n; i++) rex = max(rex, dp[n][i]);
    cout << rex;
    return 0;
}

G

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#include<iostream>
#include<algorithm>
#include<string>
#include<cmath>
using namespace std;
#define ll long long
const int N = 65;
ll n, t, dp1[N][N], dp2[N][N], shu[N][N];
string st[N];

ll gcd(ll a, ll b) {
    return b ? gcd(b, a % b) : a;
}

int main() {
    cin >> n >> t;
    getchar();
    for (int i = 1; i <= n; i++)
        getline(cin, st[i]);
    for (int i = 1; i <= n; i++) {
        int k = 1;
       // cout << st[i] << endl;
        for (int j = 0; j < st[i].size(); j++)
            if (st[i][j] == '*')  shu[i][k++] = 0;
            else if (st[i][j] == '.')  shu[i][k++] = 1;
    }
    int k = 1, r = 1;
    while (shu[k][r])  k += 2, r++;
    dp1[k][r] = 1; dp2[k][r] = 1;
    for (int i = k; i <= n; i++) {
        for (int j = 1; j <=i; j++) {
            if (i != n + 1 && shu[i][j] == 1) dp1[i + 2][j + 1] += 4 * dp1[i][j];
            else dp1[i + 1][j + 1] += dp1[i][j], dp1[i + 1][j] += dp1[i][j];
            //cout << dp1[i][j] << ' ';
        }
        //cout << endl;
    }
    ll p = 1; dp2[n+1][t+1] = pow(2, n+1 - k);
    if (dp1[n + 1][t + 1] == 0) dp2[n + 1][t + 1] = 1;
    else p = gcd(dp1[n + 1][t + 1], dp2[n + 1][t + 1]);
    cout << dp1[n + 1][t + 1] / p << '/' << dp2[n + 1][t + 1] / p ;
    return 0;
}

H

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#include<iostream>
#include<algorithm>
using namespace std;
const int N = 30;
long long qi[N][N], dp[N][N];
int n, m, x, y;
int X[8]{ 2,2,1,1,-1,-1,-2,-2 }, Y[8]{ 1,-1,2,-2,2,-2,1,-1 };

int main() {
	cin >> n >> m >> x >> y;
	n += 2, m += 2, x += 2, y += 2;
	qi[x][y] = 1;
	//cout << x << ' ' << y << endl;
	for (int i = 0; i < 8; i++)
		if (x + X[i] >= 0 && y + Y[i] >= 0)
			qi[x + X[i]][y + Y[i]] = 1;
	dp[2][2] = 1;
	for (int i = 2; i <= n; i++)
	{
		for (int j = 2; j <= m; j++) {
			if (i == 2 && j == 2||qi[i][j] == 1) continue;
			else dp[i][j] = dp[i - 1][j] + dp[i][j - 1];
			//cout << dp[i][j] << ' ';
		}
		//cout << endl;
	}
	cout << dp[n][m];
	return 0;
}

P

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#include<iostream>
#include<algorithm>
#include<algorithm>
using namespace std;
#define ll long long
const int N = 1e4 + 5;
int n, m, k, dp[105][N], v[N];

int main() {
	memset(dp, -0x3f, sizeof(dp));
	cin >> n >> m >> k;
	for (int i = 1; i <= n; i++) {
		cin >> v[i];
		dp[1][i] = max(dp[1][i - 1], v[i]);
	}
	for (int i = 2; i <= m; i++)
		for (int j = k+1; j <= n; j++)
			dp[i][j] = max(dp[i][j - 1], dp[i - 1][j-k] + v[j]);
	cout << dp[m][n];
	return 0;
}

Q

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#include<iostream>
#include<algorithm>
using namespace std;
#define ll long long
int v, n, dp[20005];
int shu[35];

int main() {
	cin >> v >> n;
	for (int i = 1; i <= n; i++)
		cin >> shu[i];
	for (int i = 1; i <= n; i++) {
		if (shu[i] > v) continue;
		for (int j = v; j >= shu[i]; j--)
			dp[j]=max(dp[j],dp[j - shu[i]] + shu[i]);
	}
	cout << v - dp[v];
	return 0;
}

R

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#include<iostream>
#include<algorithm>
#include<cstring>
using namespace std;
const int N = 1e3 + 5;
int shu[N], w[N], n, m;
int dp[N];

int main() {
    memset(dp, 0, sizeof(dp));
    cin >> n >> m;
    for (int i = 1; i <= m; i++)
        cin >> shu[i] >> w[i];
    for (int i = 1; i <= m; i++)
        for (int j = n; j >= shu[i]; j--)
            dp[j] = max(dp[j], dp[j - shu[i]] + w[i]);
    cout << dp[n];
    return 0;
}

S

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 //#include<iostream>
#include<algorithm>
using namespace std;
#define ll long long
int V, n;
int v[35],p;
ll w[35], dp[20005];

int main() {
	cin >> V >> n;
	for (int i = 1; i <= n; i++){
		cin >> v[i] >> p;
		w[i] = v[i] * p;
	}
	for (int i = 1; i <= n; i++)
		for (int j = V; j >= v[i]; j--)
			dp[j]=max(dp[j],dp[j - v[i]] + w[i]);
	cout << dp[V];
	return 0;
}

T

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#include<iostream>
#include<algorithm>
using namespace std;
#define ll long long
int v, n, dp[10005],num;
int shu[105];

int main() {
	cin  >> n;
	for (int i = 1; i <= n; i++){
		cin >> shu[i];
		num += shu[i];
	}
	v = num / 2;
	for (int i = 1; i <= n; i++)
		for (int j = v; j >= shu[i]; j--)
			dp[j]=max(dp[j],dp[j - shu[i]] + shu[i]);
	cout << min(dp[v], num - dp[v]) << ' ' << max(dp[v], num - dp[v]);
	return 0;
}