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
因为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，其实就是随便哪一列，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;
}2013 长沙邀请赛 ADEGH 题解,布布扣,bubuko.com
原文地址：http://blog.csdn.net/qq574857122/article/details/37672049
