HDU 4565 So Easy!

类似fib的构造

设Fn = x + y*sqrt(b)

啪啦啪啦

#include <cstdio>
#include <cstring>
#include <vector>
#include <cmath>
#include <iostream>

using namespace std;

typedef vector<long long> vec;
typedef vector<vec> mat;
typedef long long ll;

ll a, b, n, MOD;

mat mul(mat &A, mat &B) {
	mat C(A.size(), vec(B[0].size()));
	for (int i = 0; i < A.size(); ++i) {
		for (int k = 0; k < B.size(); ++k) {
			for (int j = 0; j < B[0].size(); ++j) {
				C[i][j] = (C[i][j] + A[i][k] * B[k][j]) % MOD;
			}
		}
	}
	return C;
}

mat pow(mat A, ll n) {
	mat B(A.size(), vec(A.size()));
	for (int i = 0; i < A.size(); ++i) {
		B[i][i] = 1;
	}
	while (n > 0) {
		if (n & 1) B = mul(B, A);
		A = mul(A, A);
		n >>= 1;
	}
	return B;
}

int main() {
	while (cin >> a >> b >> n >> MOD) {
		if (n == 1) {
			ll ans = a;
			ll bb = (ll)sqrt(1.0 * b);
			if (bb * bb != b) ++ans;
			ans = (ans + bb) % MOD;
			cout << ans << endl;
			continue;
		}
		mat A(2, vec(2));
		A[0][0] = a, A[0][1] = b;
		A[1][0] = 1, A[1][1] = a;
		A = pow(A, n - 1);
		ll ans = A[0][0] * a % MOD;
		ans = (ans + A[0][1]) % MOD;
		ans = ans * 2 % MOD;
		cout << ans << endl;
	}
	return 0;
}

状压dp，写的麻烦了。

。

#include <iostream>
#include <algorithm>
#include <cstdio>
#include <queue>
using namespace std;
#define N 205
#define M 14
#define inf 10000000
int val[N][N];
int n,m,k;
struct node{
    int x,y;
}c[M];
int d[N][N];
bool vis[N][N];
int step[4][2] = {0,1,0,-1,1,0,-1,0};
bool inmap(int x,int y){return 1<=x&&x<=n&&1<=y&&y<=m;}
void spfa(int x,int y) {
    memset(vis,0,sizeof vis);
    for(int i = 0; i <= n; i++) for(int j = 0; j <= m; j++)d[i][j] = inf;
    d[x][y] = 0;
    queue<int>qx,qy;
    qx.push(x), qy.push(y);
    while(!qx.empty()) {
        int u = qx.front(); qx.pop();
        int v = qy.front(); qy.pop();
        vis[u][v] = 0;
        for(int i = 0; i < 4; i++) {
            int nx = u+step[i][0], ny = v+step[i][1];
            if(!inmap(nx,ny))continue;
            if(d[nx][ny] > d[u][v] + val[nx][ny]) {
                d[nx][ny] = d[u][v]+val[nx][ny];
                if(vis[nx][ny]==0)
                    vis[nx][ny] = 1, qx.push(nx), qy.push(ny);
            }
        }
    }
}
void work(){
    memset(vis,0,sizeof vis);
    for(int i = 0; i <= n; i++) for(int j = 0; j <= m; j++)d[i][j] = inf;
    queue<int>qx,qy;
    for(int i = 1; i <= n; i++) {
        qx.push(i); qy.push(1);
        qx.push(i); qy.push(m);
        d[i][1] = val[i][1];
        d[i][m] = val[i][m];
    }
    for(int i = 1; i <= m; i++) {
        qx.push(1); qy.push(i);
        qx.push(n); qy.push(i);
        d[1][i] = val[1][i];
        d[n][i] = val[n][i];
    }
    while(!qx.empty()) {
        int u = qx.front(); qx.pop();
        int v = qy.front(); qy.pop();
        vis[u][v] = 0;
        for(int i = 0; i < 4; i++) {
            int nx = u+step[i][0], ny = v+step[i][1];
            if(!inmap(nx,ny))continue;
            if(d[nx][ny] > d[u][v] + val[nx][ny]) {
                d[nx][ny] = d[u][v]+val[nx][ny];
                if(vis[nx][ny]==0)
                    vis[nx][ny] = 1, qx.push(nx), qy.push(ny);
            }
        }
    }
}
int find(){
    int ans = inf;
    for(int i = 1; i <= n; i++)ans = min(ans, min(d[i][1],d[i][m]));
    for(int i = 1; i <= m; i++)ans = min(ans, min(d[1][i],d[n][i]));
    return ans;
}
int dis[20][20];
int dp[1<<M][M];
bool can[M];
int main(){
    int i,j,T;scanf("%d",&T);
    while(T--){
        scanf("%d %d",&n,&m);
        for(i = 1; i <= n; i++)
            for(j = 1; j <= m; j++) {
                scanf("%d", &val[i][j]);
                if(val[i][j]<0)val[i][j] = inf;
            }
        scanf("%d",&k);
        for(i = 0; i < k; i++) {scanf("%d %d",&c[i].x,&c[i].y); c[i].x++; c[i].y++;}
        c[k].x = c[k].y = 0;
        for(i = 0; i < k; i++) {
            spfa(c[i].x, c[i].y);
            for(int j = 0; j < k; j++)
            {
                dis[i][j] = d[c[j].x][c[j].y];
            }
            dis[i][k] = find();
        }
        work();
        for(i = 0; i < k; i++)
            dis[k][i] = d[c[i].x][c[i].y];
        bool nothing = true;
        for(i = 0; i < k; i++) {
            can[i] = (d[c[i].x][c[i].y])<inf;
            if(can[i])nothing = false;
        }
        if(nothing){ puts("0"); continue;  }
        int all = (1<<k)-1;
        for(i = 0; i <= all; i++) for(j = 0; j < k; j++)    dp[i][j] = inf;
        for(i = 0; i < k; i++) if(can[i])dp[1<<i][i] = dis[k][i];
        for(i = 0; i <= all; i++) {
            for(j = 0; j < k; j++) if(dp[i][j]<inf && can[j]) {
                for(int K = 0; K < k; K++)
                    if((i&(1<<K))==0 && can[K]) {
                        dp[i|(1<<K)][K] = min(dp[i][j]+dis[j][K], dp[i|(1<<K)][K]);
                    }
            }
        }
        int ans = inf;
        all = 0;
        for(i = 0; i < k; i++)
            if(can[i]) all |= (1<<i);
        for(i = 0; i < k; i++) if(can[i])
            ans = min(ans, dp[all][i] + dis[i][k]);
        cout<<ans<<endl;
    }
    return 0;
}

HDU 4569 Special equations

找循环节

但p*p太大

由于f(x) % (p*p) ==0

所以f(x) % p ==0

所以先找到%p=0的第一个解 x

则让x += p 再找循环节，若找到了循环节还无解，则就是无解了

#include <cstdio>
#include <algorithm>
#include <cmath>
#include <iostream>
typedef long long ll;
const int N = 100000000;

int cas = 0;
int a[10], n;

int get(int x, int mod) {
    int re = 0, k = 1;
    for (int i = 0; i <= n; ++i) {
        re = ((ll)re + (ll)k *a[i]) % mod;
        k = (ll)k * x % mod;
    }
    return re;
}

void work() {
    ++ cas;
    int m, up;
    scanf("%d", &n);
    for (int i = n;  i >= 0; --i)
        scanf("%d", &a[i]);
    scanf("%d", &m);
    up = m * m;

    int x, ans = -1, g;
    for (int i = 0; i <= m; ++i) {
        g = 0;
        x = i;
        for (int j = 1; j <= n; ++j) {
            g = ((ll)g + (ll)a[j] * x) % m;
            x = ((ll)x * i) % m;
        }
        if (g == ((-a[0]) % m + m) % m) {
            //printf("%d\n", i);
            for (int j = i; j <= up; j += m) {
                if (get(j, up) == 0) {
                    ans = j;
                    break;
                }
            }
        }
        if (ans >= 0)
            break;
    }

    printf("Case #%d: ", cas);
    if (ans == -1)
        puts("No solution!");
    else
        printf("%d\n", ans);
}

int main() {
    int t;  scanf("%d", &t);
    while (t--)       work();
    return 0;
}

01背包。先排个序这样就能不考虑递增的条件了

#include <iostream>
#include <algorithm>
#include <cstdio>
using namespace std;
const int Inf = (int)(1e9) + 10;
const int N = 100 + 2;
const int T = 300 + 2;

int n, s, t, lim, tcas = 0;
int d[T][N], w[N][N], cos[N], val[N];

int dp(int cur, int u) {
    if (~d[cur][u])
        return d[cur][u];
    d[cur][u] = 0;
    for (int i = 0; i < n; ++i)
        if (val[i] > val[u] && cur + w[u][i] + cos[i] + w[i][t] <= lim) {
            d[cur][u] = max(d[cur][u], dp(cur + w[u][i] + cos[i], i) + val[i]);
        }
    return d[cur][u];
}

void work() {
    int m, u, v, len;
    memset(d, -1, sizeof d);
    scanf("%d%d%d%d%d", &n, &m, &lim, &s, &t);
    for (int i = 0; i < n; ++i)
        scanf("%d", &cos[i]);
    for (int i = 0; i < n; ++i)
        scanf("%d", &val[i]);

    for (int i = 0; i <= n; ++i) {
        for (int j = 0; j <= n; ++j)
            w[i][j] = Inf;
        w[i][i] = 0;
    }
    while (m -- > 0) {
        scanf("%d%d%d" ,&u, &v, &len);
        if (w[u][v] > len) {
            w[u][v] = len;
            w[v][u] = len;
        }
    }
    w[n][s] = w[s][n] = 0;

    for(int k = 0; k <= n; ++k)
        for (int i = 0; i <= n; ++i)
        for (int j = 0; j <= n; ++j)
            if (w[i][j] > w[i][k] +w[k][j])
                w[i][j] = w[i][k] +w[k][j];


    cos[n] = 0;
    val[n] = -1;

    printf("Case #%d:\n", ++tcas);
    if(w[n][t] > lim)
        puts("0");
    else
        printf("%d\n", dp(0, n));
}

int main() {
    int cas;
    scanf("%d", &cas);
    while (cas -- > 0)
        work();
    return 0;
}

HDU 4572 Bottles Arrangement

先构造一个矩阵

11223344555544332211

然后后移一位构造出第二行

然后找个规律。事实上就是随便哪一列，反正答案都是一样的

交换行和交换列操作：

交换行操作不影响结果。交换列操作是非法的。。

#include <cstdio>
int main() {
    int n, m;
    while(scanf("%d%d", &n, &m)!=EOF) {
        int ans = n + n - 1;
        int x = n + 1, y = n - 1;
        for(int i = 3; i <= m; i ++) {
            if(i&1) {
                ans += --x;
            } else ans += --y;
        }
        printf("%d\n", ans);
    }
    return 0;
}