Provincial competition exercise 2 -- the 8th Fujian college student programming competition & supplementary questions

              Let's assume that there is already a A card , If we want to get the second a+1 Zhang , The probability of our winning it is obviously (n-a)/n, The reciprocal of probability is its expectation , So his expectation is n/(n-a), If you add up, the total expectation is n(1+1/2+1/3+...+1/n), And then multiplied by the （n-1）!, The final formula is n！(1+1/2+1/3+...+1/n). Try to avoid two different types of large number operations .

import java.math.BigDecimal;
import java.math.BigInteger;
import java.util.Scanner;

public class YYS {
	static  BigInteger jie(int x)
	{
		BigInteger sum= BigInteger.ONE;
	    for(int i=1;i<=x;i++)
	    {
	    	BigInteger z= BigInteger.valueOf(i);
	        sum=sum.multiply(z);
	    }
	    return sum;
	}
	/*static BigDecimal qi(int x)
	{
		BigDecimal sum =BigDecimal.valueOf(0);
	    for(double i=1;i<=x;i++)
	    {
	    	double z=1/i;
	        sum=sum.add(BigDecimal.valueOf(z));
	    }
	    return sum.multiply(BigDecimal.valueOf(x));
	}
*/
	public static void main(String[] args) {
	    int t=0;;
	    Scanner cin=new Scanner(System.in);
	    t=cin.nextInt();
	    for(int k=0;k<t;k++)
	    {
	    	
	        int n;
	        n=cin.nextInt();
	        BigInteger ans=BigInteger.ZERO;
	        BigInteger sum1=jie(n);
        
            for(int i=1;i<=n;i++){
                ans=ans.add(sum1.divide(BigInteger.valueOf(i)));
            }
            System.out.println(ans+".0");
	       /* BigDecimal sum1=jie(n-1);
	        BigDecimal sum2=qi(n);
	        BigDecimal aaa=sum1.multiply(sum2);
	        System.out.println(String.format("%.1f", aaa));*/
	    }


	}

}

The question ：

business GTX1920, We know that every piece in the future GTX1920 The price of , Ask about the biggest profit after trading

• Buy as much as you can every time you buy , Sell it all every time
• Buy in every local price Valley , Then sell at the top of the first local price
• The final funds will be large , exceed long long The scope of the , But the intermediate result cannot be modeled , Otherwise, you can't calculate the next purchase GTX1920 The number of , So use BigInteger To calculate

import java.util.Scanner;
import java.math.BigInteger;

public class Main {
    public static void main(String[] args) {
        Scanner cin=new Scanner(System.in);
        int T;
        int n,m;
        int[] c = new int[2005];
        T=cin.nextInt();
        BigInteger ans,mod=BigInteger.valueOf(1000000007);
        for(int xx=1;xx<=T;++xx){
            n=cin.nextInt();
            m=cin.nextInt();
            ans=BigInteger.valueOf(m);
            for(int j=0;j<n;++j) {
                c[j]=cin.nextInt();
            }
            int i,j,k;
            for(i=0;i<n;) {
                j=i;
                while(j<n-1&&c[j+1]<=c[j])++j;
                if(j==n-1)break;
                k=j;
                while(k<n-1&&c[k+1]>c[k])++k;
                BigInteger res=ans.mod(BigInteger.valueOf(c[j]));
                BigInteger buy=ans.divide(BigInteger.valueOf(c[j]));
                buy=buy.multiply(BigInteger.valueOf(c[k]));
                ans=res.add(buy);
                i=k+1;
            }
            System.out.println("Case #"+xx+": "+ans.mod(mod));
        }
    }
}