ACM-ICPC 2018 焦作赛区网络预赛 题解
A
#includeelse
cout<<"Dare you say that again?"< B
dp[i][j][0]表示使用了前i个符号,以第j个数字结尾 的最大值
dp[i][j][1]表示使用了前i个符号,以第j个数字结尾 的最小值
#includescanf("%lld",&a[i]);
}
getchar();
scanf("%s",s);
len=strlen(s);
for(int i=0;ifor(int j=0;j0]=-inf;
dp[i][j][1]=inf;
}
maxx=k;
minn=k;
int tj=0;
for(int i=0;iif(i)
{
while(dp[i-1][tj][0]==-inf)
tj++;
maxx=dp[i-1][tj][0];
minn=dp[i-1][tj][1];
}
for(int j=i?tj+1:0;jif(s[i]=='/'&&a[j]==0)
{
if(i&&dp[i-1][j][0]!=-inf)
{
maxx=max(maxx,dp[i-1][j][0]);
minn=min(minn,dp[i-1][j][1]);
}
continue;
}
ans1=cal(maxx,s[i],a[j]);
ans2=cal(minn,s[i],a[j]);
dp[i][j][0]=max(ans1,ans2);
dp[i][j][1]=min(ans1,ans2);
if(i&&dp[i-1][j][0]!=-inf)
{
maxx=max(maxx,dp[i-1][j][0]);
minn=min(minn,dp[i-1][j][1]);
}
}
}
maxx=-inf;
for(int j=(len!=1)?tj+1:0;j1][j][0]);
printf("%lld\n",maxx);
}
} C
D
E
F
G
指数循环节+快速幂
#include10+n[i]-'0';
ans--;
}
else
{
for(int i=0;i10+n[i]-'0';
ans%=p;
}
ans--;
ans+=p;
ans%=p;
ans+=p;
}
printf("%lld\n",poww(2,ans));
}
} H
后缀自动机（SAM）模板题
I
import java.util.*;
public class Main {
static Scanner sc = new Scanner(System.in);
public static void main(String[] args) {
while(sc.hasNext()){
int a = sc.nextInt(), b = sc.nextInt(), c = sc.nextInt();
if(a*b*c%2==0)
System.out.println("Yes");
else
System.out.println("No");
}
}
}J
套个JAVA大数开方的板子就OK了
import java.math.BigInteger;
import java.util.Scanner;
public class Main {
public static void main(String[] args) {
Scanner cin=new Scanner(System.in);
int t=cin.nextInt();
while(t-->0) {
String str=cin.next();
BigInteger n=new BigInteger(str);
BigInteger m=n.multiply(n.subtract(BigInteger.ONE)).shiftRight(1);
boolean ni=isSquare(n),mi=isSquare(m);
if(ni&&mi)
System.out.println("Arena of Valor");
else if(ni&&!mi)
System.out.println("Hearth Stone");
else if(!ni&&mi)
System.out.println("Clash Royale");
else
System.out.println("League of Legends");
}
cin.close();
}
public static boolean isSquare(BigInteger n) {
BigInteger l=BigInteger.ZERO,r=n,mid,ans;
int res;
while(l.compareTo(r)<=0) {
mid=l.add(r).shiftRight(1);
ans=mid.multiply(mid);
res=ans.compareTo(n);
if(res==0)
return true;
else if(res<0)
l=mid.add(BigInteger.ONE);
else
r=mid.subtract(BigInteger.ONE);
}
return false;
}
}
K
分组背包的简单变形
#include1<for(int j=10000;j>=k;j--)
{
dp[j]+=dp[j-k];
dp[j]%=mo;
}
}
}
while(q--)
{
scanf("%d",&s);
printf("%d\n",dp[s]);
}
}
return 0;
} L
矩阵快速幂，由数位DP的思想很容易就能找出递推式
#include