Shen Huang's Blog
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Posted on Edited on

很容易想到类似素数筛选的方法，但是发现空间这个题目开的很小，所以开那么大空间不行

发现d(n) - n < 7 * 9，函数与自身相减是小于63,某个数，最多只会影响它的第63个后面的数，考虑用一个类似滚动数组的hash，如果这个数是，那么就存储，如果这个数不是。就不做

不管怎样都要把后面一个消除影响以及这个数还原为可以的

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/* From: Lich_Amnesia
 * Time: 2013-12-20 22:34:03
 *
 * SGU 108
 * */
#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstring>
#include <cmath>
#include <queue>
#include <set>
#include <vector>
using namespace std;

const int INF = ~0u>>1;
typedef pair <int,int> P;
#define MID(x,y) ((x+y)>>1)
#define iabs(x) ((x)>0?(x):-(x))
#define REP(i,a,b) for(int i=(a);i<(b);i++)
#define FOR(i,a,b) for(int i=(a);i<=(b);i++)
#define pb push_back
#define mp make_pair
#define print() cout<<"--------"<<endl
#define maxn 5010

struct node{
	int id,ans,val;
}num[maxn];

bool cmp(node a,node b){
	return a.val < b.val;
}

int val[10010];
bool vis[66];
int ans[maxn];
int main(){
	int n,k;
	scanf("%d%d", &n, &k);
	for (int i = 0; i < k; i++){
		scanf("%d", &num[i].val);
		num[i].id = i;
	}
	int len = 0;
	int now = 0;
	sort(num,num + k,cmp);

	//求d(n)的数字部分的和
	for (int i = 1; i <= 10000; i ++){
		val[i] = val[i /10] + i % 10;
	}
	memset(vis,1,sizeof(vis));

	for (int i = 1; i <= n; i ++){
		if (vis[i % 63]){
			len ++;
			while (num[now].val == len){
				ans[num[now++].id] = i;
			}
		}

		vis[i % 63] = 1;

		int next = i + val[i / 10000] + val[i % 10000];
		vis[next % 63] = 0;

	}
	cout << len << endl;
	for (int i = 0; i < now; i++){
		if (i) printf(" ");
		printf("%d", ans[i]);
	}
	cout << endl;
	return 0;
}

Posted on Edited on 
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/* From: Lich_Amnesia
 * Time: 2013-12-16 20:21:11
 *
 * POJ 3420 DP + 状态压缩 + 矩阵快速幂
 * */
#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstring>
#include <cmath>
#include <queue>
#include <set>
#include <vector>
using namespace std;

const int INF = ~0u>>1;
typedef pair <int,int> P;
#define MID(x,y) ((x+y)>>1)
#define iabs(x) ((x)>0?(x):-(x))
#define REP(i,a,b) for(int i=(a);i<(b);i++)
#define FOR(i,a,b) for(int i=(a);i<=(b);i++)
#define pb push_back
#define mp make_pair
#define print() cout<<"--------"<<endl
typedef long long ll;
#define maxn 100010
ll ans[16][16];
#define N 16
void product_mod(ll a[][16], ll b[][16],int n, ll mod){
	ll c[N][N] = {};
	for (int i = 0; i < n; i ++){
		for (int j = 0; j < n; j ++){
			for (int k = 0; k < n; k ++){
				c[i][j] += a[i][k] * b[k][j];
			}
			c[i][j] %= mod;
		}
	}
	for (int i = 0; i < n; i++){
		for (int j = 0; j < n; j++){
			a[i][j] = c[i][j];
		}
	}
}

void power_mod(ll a[][N], ll x, int n, ll mod){
	memset(ans, 0, sizeof(ans));
	for (int i = 0; i < n; i ++) ans[i][i] = 1;
	while (x){
		if (x&1) product_mod(ans, a, n, mod);
		product_mod(a, a, n, mod);
		x >>= 1;
	}
}
ll a[N][N];
int bits[N];

void init(){
	for (int i = 0; i < 4; i++)
		bits[i] = (1<<i);
	memset(a,0,sizeof(a));
	int tmp;
	for (int i = 0; i < 16; i++){
		a[i][(~i & 0xf)] = 1;	
		tmp = (~i & 0xf);
		for (int j = 0; j < 3; j ++){
			if ((tmp & bits[j]) == 0 && (tmp & bits[j + 1]) == 0){
				a[i][tmp | bits[j] | bits[j+1]] = 1;
			}
		}
	}
	a[15][15] = 1;
	/*for (int i = 0; i < 16; i++){
		for (int j = 0; j < 16; j++)
			cout << a[i][j] << &#39; &#39; ;
		cout << endl;
	}*/
}

int main(){
	int n, mod;
	while (~scanf("%d%d", &n, &mod) && (n + mod)){
		init();
		power_mod(a, n, 16, (long long)mod);
		printf("%lldn", ans[15][15]);
	}
	return 0;
}

用矩阵表示状态的转移

位运算计算下一行的状态

两个能够转化的连接一条边

Posted on Edited on

dp[i][x][y] 表示前i个已经完成匹配并且i+1为x，i+2为y的时候的最小变化次数

因为第i个是肯定要变的，然后后面的两个数的变换次数分别有up >= k >= l （up为i的变化，k为i+1的变换，l为i+2的变换）或者down >= k >= l

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/* From: Lich_Amnesia
 * Time: 2013-12-16 19:14:28
 *
 * HDU 4433 DP
 * */
#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstring>
#include <cmath>
#include <queue>
#include <set>
#include <vector>
using namespace std;

const int INF = 100000000;
typedef pair <int,int> P;
#define MID(x,y) ((x+y)>>1)
#define iabs(x) ((x)>0?(x):-(x))
#define REP(i,a,b) for(int i=(a);i<(b);i++)
#define FOR(i,a,b) for(int i=(a);i<=(b);i++)
#define pb push_back
#define mp make_pair
#define print() cout<<"--------"<<endl
#define maxn 1010

char s[maxn],t[maxn];
int dp[maxn][12][12];
int a[maxn],b[maxn];
int main(){
	while (~scanf("%s%s", s, t)){
		for (int i = 0; i <= strlen(s); i ++) 
			for (int j = 0; j < 10; j ++)
				for (int l = 0; l < 10; l ++)
					dp[i][j][l] = INF;

		for (int i = 0; i < strlen(s); i++){
			a[i+1] = s[i] - &#39;0&#39;;
			b[i+1] = t[i] - &#39;0&#39;;
		}
		int len = strlen(s);
		a[len + 1] = a[len + 2] = 0;
		b[len + 1] = b[len + 2] = 0;

		dp[0][a[1]][a[2]] = 0;

		for (int i = 1; i <= len; i ++){
			for (int x = 0; x < 10; x++){
				for (int y = 0; y < 10; y++){
					int down = (x + 10 - b[i]) % 10;
					for (int k = 0; k <= down; k++)
						for (int l = 0; l <= k; l++){
							dp[i][(y-k+10)%10][(a[i+2]-l+10)%10] = 
							min(dp[i-1][x][y] + down,
									dp[i][(y-k+10)%10][(a[i+2]-l+10)%10]);
						}
					int up = 10 - down;
					for (int k = 0; k <= up; k++)
						for (int l = 0; l <= k; l++){
							dp[i][(y+k)%10][(a[i+2]+l)%10] =
							min(dp[i-1][x][y] + up,dp[i][(y+k)%10][(a[i+2]+l)%10]);
						}
				}
			}
		}
		printf("%dn", dp[len][0][0]);
	}
	return 0;
}

Posted on Edited on

画图找规律

f[1] = 2
f[n] = f[n-1] + 1 + f[n-1] + 1 + f[n-1]
f[n] = 3^n - 1 快速幂

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#include <iostream>
#include <cmath>
#include <cstdio>

using namespace std;
typedef long long ll;

ll quick_mod(ll n, ll k, ll mod){
	ll ret = 1;
	while (k){
		if (k&1) ret = (ret * n) % mod;
		k >>= 1;
		n = (n*n) % mod;
	}
	return ret;
}

int main(){
	ll n,m;
	cin >> n >> m;
	cout << (quick_mod(3,n,m) - 1 + m) % m<< endl;
 	return 0;
}

Posted on Edited on

做一个映射表第一次出现的位置就行了

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#include <iostream>
#include <cmath>
#include <algorithm>
#include <map>
#include <cstdio>
using namespace std;
typedef long long ll;
#define maxn 100010
int a[maxn],b[maxn];
map<int,int>map1,map2;
bool cmp(int a,int b){
	return a > b;
}
int main(){
	int n,m;
	cin >> n;
	for (int i = 0; i < n; i++)
	{	
		cin >> a[i];
		b[n - i - 1] = a[i];
	}
	int num = 1;
	for (int i = 0; i < n; i++){
		if (map1[a[i]] == 0) map1[a[i]] = i + 1;
	}

	for (int i = 0; i < n; i++){
		if (map2[b[i]] == 0) map2[b[i]] = i + 1;
	}
	cin >> m;
	int cnt;
	ll ansmin = 0,ansmax = 0;
	for (int i = 0; i < m; i++){
		cin >> cnt;
		if (map1[cnt] == 0) ansmin += n;
		else ansmin += map1[cnt];
		if (map2[cnt] == 0) ansmax += n;
		else ansmax += map2[cnt];
	}
	cout << ansmin << &#39; &#39; << ansmax << endl;
 	return 0;
}

Posted on Edited on 
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#include <iostream>
#include <cmath>
#include <cstdio>

using namespace std;
typedef long long ll;
ll x[10],y[10];

int main(){
	cin >> x[0] >> y[0];
	cin >> x[1] >> y[1];
	cin >> x[2] >> y[2];
	x[3] = x[1] - x[0];y[3] = y[1] - y[0];
	x[4] = x[2] - x[1];y[4] = y[2] - y[1];
	ll ans = x[3] * y[4] - x[4] * y[3];
	if (ans > 0) printf("LEFTn");
	else if (ans < 0) printf("RIGHTn");
 	else printf("TOWARDSn");
 	return 0;
}

Posted on Edited on

想象成一个树，然后有一个递归关系,

1.如果这个点src的子树中有一个含有2号边那就不用管直接return 1表示这个以src为根的子树已经判定完毕

2.如果这个点src正好是叶子节点那么就返回[src][pre]这条边的值，1表示需要修理，并且存下这个src必须得需要它这个点

3.如果这个src的子树中没有含有2号点的证明src下面的树中没有需要修理的。

两种情况：第一个是src和pre这条边是2号边那就return 1，表示这个点src是答案中的。第二个是如果src和pre的是1号边那么就return 0,不需要修理

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#include <iostream>
#include <algorithm>
#include <map>
#include <cstdio>
#include <set>
#include <cstring>
#include <vector>
using namespace std;

struct edge{
	int u,t;
}tmp;

#define maxn 100010
vector <edge> ve[maxn];
bool vis[maxn];
int ans[maxn],ansnum = 0;

bool solve(int src, int pre, int flag){
	vis[src] = 1;
	int cnt = 0;
	int ret = -1;
	int sumret = 0;
	for (int i = 0; i < ve[src].size(); i ++){
		//if (ve[src][i].u == pre) yeziret = ve[src][i].t ;
		if (!vis[ve[src][i].u]){
			ret = solve(ve[src][i].u,src,ve[src][i].t);
			sumret += ret;
		}
		else cnt ++;
		if (i == ve[src].size() - 1 && sumret == 0 && flag == 1 && cnt != ve[src].size()){
			ans[ansnum ++] = src;
			return flag;
		}
		if (i == ve[src].size() - 1 && cnt != ve[src].size()) return sumret;
	}
	if (cnt == ve[src].size()){
		if (flag) ans[ansnum ++] = src;
		return flag;
	}
}

int main(int argc, char const *argv[])
{
	int n,u,v,t;
	memset(vis,0,sizeof(vis));
	scanf("%d", &n);
	for (int i = 0; i < n - 1; i++){
		scanf("%d%d%d", &u, &v, &t);
		tmp.u = v;
		tmp.t = t - 1;
		ve[u].push_back(tmp);
		tmp.u = u;
		tmp.t = t - 1;
		ve[v].push_back(tmp);
	}
	solve(1,1,0);
	printf("%dn", ansnum);
	for (int i = 0; i < ansnum; i++)
		printf("%d ", ans[i]);
	return 0;
}

Posted on Edited on

分成k和n-k的部分，然后算sk/k 和(sall - sk) /(n - k)，易知，前面的k个数肯定是比较大的，然后前面k个进行均分sk%k的值在前k中分配且要不能大于r，然后后面的n-k也是一样的只是不能大于a[k]

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#include <iostream>
#include <algorithm>
#include <map>
#include <cstdio>
using namespace std;
int a[1010];
int main(int argc, char const *argv[])
{
	int n,k,l,r,sall,sk;
	scanf("%d%d%d%d%d%d", &n, &k, &l, &r, &sall, &sk);
	int Maxn = sk / k;
	int Maxmod = sk % k;
	for (int i = 0; i < k; i++){
		a[i] = Maxn;
		if (Maxmod){
			if (Maxmod + a[i] >= r) {
				Maxmod -= r - a[i];
				a[i] = r;
			}else {
				a[i] += Maxmod;
				Maxmod = 0;
			}
		}
	}
	if (n > k){
		Maxn = (sall - sk) / (n-k);
		Maxmod = (sall - sk) % (n-k);

		for (int i = k; i < n; i++){
			a[i] = Maxn;
			if (Maxmod){
				if (Maxmod + a[i] >= a[k-1]) {
					Maxmod -= a[k-1] - a[i];
					a[i] = a[k-1];
				}else {
					a[i] += Maxmod;
					Maxmod = 0;
				}
			}
		}
	}
	for (int i = 0; i < n; i ++){
		printf("%d ", a[i]);
	}

	return 0;
}

Posted on Edited on

判定1，2。算一下 有没有可以用的盘子或者碗，然后加减一下就行

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#include <iostream>
#include <algorithm>
#include <map>
#include <cstdio>
using namespace std;
int a[1010];
int main(int argc, char const *argv[])
{
	int n,m,k;
	scanf("%d%d%d", &n, &m, &k);
	int bow = m, plat = k;
	int ans = 0;
	for (int i = 0; i < n; i ++){
		scanf("%d", &a[i]);
		if (a[i] == 1) {
			if (bow) bow --;
			else ans ++;
		}else {
			if (plat) {
				plat --;
			}else if (bow){
				bow --;
			}else ans ++;
		}
	}
	cout << ans << endl;

	return 0;
}

Posted on Edited on

找出规律，最后一位对数据的影响只在和最前一位的大小关系上，然后去掉最后一位其他的都是可行的，最后加上1-9。