EXAM-2018-8-9
EXAM-2018-8-9
B
水题 注意理解题意 有坑
G
博弈 观察发现 总是会进行到最后两个,或者先手取完全部,再特判一下只有一张牌的情况
H
九连环 通过找规律 我们可以得出递推式:
F[n]=F[n-1]+2*F[n-2]+1
而这个递推式可以通过构造矩阵,然后矩阵快速幂解决:
|F[n-1],F[n-2],1| * A = |F[n],F[n-1],1|
A=
1 1 0
2 0 0
1 0 1
初始矩阵为|2 1 1|
package hgf;
import java.io.BufferedInputStream;
import java.io.BufferedReader;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.math.BigDecimal;
import java.math.BigInteger;
import java.util.Arrays;
import java.util.Scanner;
import java.util.StringTokenizer;
class InputReader {
BufferedReader buf;
StringTokenizer tok;
InputReader() {
buf = new BufferedReader(new InputStreamReader(System.in));
}
boolean hasNext() {
while (tok == null || !tok.hasMoreElements()) {
try {
tok = new StringTokenizer(buf.readLine());
} catch (Exception e) {
return false;
}
}
return true;
}
String next() {
if (hasNext())
return tok.nextToken();
return null;
}
int nextInt() {
return Integer.parseInt(next());
}
long nextLong() {
return Long.parseLong(next());
}
double nextDouble() {
return Double.parseDouble(next());
}
BigInteger nextBigInteger() {
return new BigInteger(next());
}
BigDecimal nextBigDecimal() {
return new BigDecimal(next());
}
}
public class Main {
static InputReader cin=new InputReader();
static PrintWriter cout=new PrintWriter(System.out);
public static void main(String[] args) {
int time,k;
time=cin.nextInt();
BigInteger arr[]=new BigInteger [3];
BigInteger f[][]=new BigInteger [3][3];
for(int z=1;z<=time;z++){
k=cin.nextInt();
if(k==1){
System.out.println(1);
continue;
}
if(k==2){
System.out.println(2);
continue;
}
arr[0]=B(2);arr[1]=B(1);
arr[2]=B(1);
f[0][0]=B(1);f[0][1]=B(1);f[0][2]=B(0);
f[1][0]=B(2);f[1][1]=B(0);f[1][2]=B(0);
f[2][0]=B(1);f[2][1]=B(0);f[2][2]=B(1);
k=k-2;
BigInteger c[]=new BigInteger [3];
BigInteger d[][]=new BigInteger [3][3];
for(;k!=0;k>>=1){
if(k%2==1){
Arrays.fill(c, B(0));
for(int i = 0; i < 3; i++) {
for(int j = 0; j < 3; j++) {
c[i] = c[i].add(arr[j].multiply(f[j][i]));
}
}
for(int i = 0; i < 3; i++) {
arr[i] = c[i];
}
}
for(int i = 0; i < 3; i++) {
for(int j = 0; j < 3; j++) {
d[i][j]=B(0);
}
}
for(int i = 0; i < 3; i++) {
for(int j = 0; j < 3; j++) {
for(int p = 0; p < 3; p++) {
d[i][j] = d[i][j].add(f[i][p].multiply(f[p][j])) ;
}
}
}
for(int i = 0; i < 3; i++) {
for(int j = 0; j < 3; j++) {
f[i][j] = d[i][j];
}
}
}
System.out.println(arr[0]);
}
cout.close();
}
static BigInteger B(int x) {
return BigInteger.valueOf(x);
}
}
BigInteger 申请1e5会超时
M
最讨厌看到有两个因素影响的题.....
题意是有l件衣服,m件洗衣机,n件烘干机,然后每件每次只能工作一件。然后这题是贪心做,最慢洗的配最快烘干的,然后肯定这几个机器是同时工作的,用d[i]维护一下。
单调队列的写法,注意一下。
#include <bits/stdc++.h>
#define ll long long
using namespace std;
const int maxn=1e5+7;
const int maxm=1e6+7;
ll l,n,m,w[maxn],d[maxn],t[maxm];
template<class T>
void read(T &res)
{
res = 0;
char c = getchar();
T f = 1;
while(c < '0' || c > '9')
{
if(c == '-') f = -1;
c = getchar();
}
while(c >= '0' && c <= '9')
{
res = res * 10 + c - '0';
c = getchar();
}
res *= f;
}
template<class T>
void out(T x)
{
if(x < 0)
{
putchar('-');
x = -x;
}
if(x >= 10)
{
out(x / 10);
}
putchar('0' + x % 10);
}
struct node{ll id;ll t;}u,v;
bool operator<(node a,node b){return a.t>b.t;}
priority_queue<node>x,h;
int main()
{
read(l);
read(n);
read(m);
for(int i=1;i<=n;i++){
read(w[i]);
u.id=i;
u.t=w[i];
x.push(u);
}
for(int i=1;i<=m;i++){
read(d[i]);
v.id=i;
v.t=d[i];
h.push(v);
}
for(int i=1;i<=l;i++){
node q=x.top();
x.pop();
t[i]=q.t;
ll ab=q.id;
u.t=q.t+w[ab];
u.id=q.id;
x.push(u);
}
ll ans=0;
for(int i=l;i>=0;i--){
node q=h.top();
h.pop();
t[i]+=q.t;
ll ab=q.id;
v.t=q.t+d[ab];
v.id=q.id;
h.push(v);
ans=max(ans,t[i]);
}
out(ans);
return 0;
}
A
应该要特意总结一下的莫队算法。模板题。先分成sqtr(n)块,然后按块排序,而不是单纯先排左端点再排右端点。然后两个指针跳来跳去的看似暴力却不是暴力的算法。
#include <bits/stdc++.h>
#define ll long long
using namespace std;
const int maxn=1e5+10;
ll ans[maxn],ant;
int block,m,n,val[maxn],num[maxn],k;
struct node
{
int l,r,id;
inline bool operator <(node cmp) const {
if(l/block!=cmp.l/block)
return l/block<cmp.l/block;
return r/block<cmp.r/block;
}
}a[maxn];
void init()
{
for(int i=1;i<=n;i++){
scanf("%d",&val[i]);
val[i]^=val[i-1];
}
for(int i=1;i<=m;i++){
scanf("%d %d",&a[i].l,&a[i].r);
a[i].l--;
a[i].id=i;
}
}
inline void add(int i)
{
ant+=num[val[i]^k];
num[val[i]]++;
}
inline void remov(int i)
{
num[val[i]]--;
ant-=num[val[i]^k];
}
int main()
{
scanf("%d %d %d",&n,&m,&k);
block=sqrt(n);
init();
sort(a+1,a+n+1);
ll l=1,r=0;
for(int i=1;i<=m;i++){
while(l>a[i].l){
l--;
add(l);
}
while(l<a[i].l){
remov(l);
l++;
}
while(r>a[i].r){
remov(r);
r--;
}
while(r<a[i].r){
r++;
add(r);
}
ans[a[i].id]=ant;
}
for(int i=1;i<=m;i++)
{
printf("%lld\n",ans[i]);
}
return 0;
}
E
北上广深算法?233333
裸模板题
#include <bits/stdc++.h>
#define ll long long
using namespace std;
const int mod=1e9+7;
ll g,p,m,a,b,n;
map <ll,ll> mp;
template<class T>
void read(T &res)
{
res = 0;
char c = getchar();
T f = 1;
while(c < '0' || c > '9')
{
if(c == '-') f = -1;
c = getchar();
}
while(c >= '0' && c <= '9')
{
res = res * 10 + c - '0';
c = getchar();
}
res *= f;
}
template<class T>
void out(T x)
{
if(x < 0)
{
putchar('-');
x = -x;
}
if(x >= 10)
{
out(x / 10);
}
putchar('0' + x % 10);
}
int qpow(ll x,ll k)
{
ll ans=1;
while(k>0)
{
if(k&1)
ans=(ans*x)%p;
x=(x*x)%p;
k>>=1;
}
return ans;
}
ll bsgs(ll x)
{
ll j=0,cnt=1;
for(; j<=m; ++j)
{
if(mp[(cnt*a)%p])
return mp[(cnt*a)%p]-j;
cnt=(cnt*g)%p;
}
return 0;
}
int main()
{
ll i;
read(g);
read(p);
m=ceil(sqrt(p));
ll cnt=qpow(g,m),ans=cnt;
mp[ans]=m;
for(i=2;i<=m;++i)
{
ans=(ans*cnt)%p;
mp[ans]=i*m;
}
read(n);
while(n--)
{
read(a);
read(b);
out(qpow(b,bsgs(a)));
printf("\n");
}
return 0;
}
不要忘记努力,不要辜负自己
欢迎指正 QQ:1468580561