URAL1036Lucky Tickets（DP+高精度）

最新推荐文章于 2018-09-26 23:16:01 发布

mowayao

最新推荐文章于 2018-09-26 23:16:01 发布

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分类专栏： DP 文章标签： URAL

本文链接：https://blog.youkuaiyun.com/mowayao/article/details/15337131

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本文介绍了一种使用动态规划（DP）解决幸运数字问题的算法。幸运数字是指其前一半数字之和等于后一半数字之和的数字。通过分析问题并构建合适的DP状态转移方程，实现快速计算所有符合条件的数字数量。此外，文章还讨论了高精度计算的必要性，以确保答案的准确性。

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You are given a number 1 ≤ N ≤ 50. Every ticket has its 2N-digit number. We call a ticket lucky, if the sum of its first N digits is equal to the sum of its last N digits. You are also given the sum of ALL digits in the number. Your task is to count an amount of lucky numbers, having the specified sum of ALL digits.

Input

Two space-separated numbers: N and S. Here S is the sum of all digits. Assume that 0 ≤ S ≤ 1000.

Output

The amount of lucky tickets.

Sample

input	output
2 2	4

Hint

The tickets are 0101, 0110, 1001, 1010 in the example above

一开始DP公式推错了(dp(i,j)=dp(i-1,j)+dp(i,j-1))，和之前ZOJ3216有点象，但是有很大的区别：这道题是10进制，每一位的大小只能取0-9而那一题能取0-n;

最终的DP公式为dp(i,j) = dp(i,j)+dp(i-1,j-k）0<=k<=9 i为数字的位数 j为和的大小

这样就结束了吗？？？高精度啊！！！！！！

#include <iostream>
#include <cstdio>
#include <cstring>
#include <vector>
#include <algorithm>
#include <string>
#include <queue>
using namespace std;
const int maxn = 100+10;
const int ten[4] = {1,10,100,1000};
struct BigNum{
    int d[maxn];
    BigNum(string s){
        int len =s.size();
        d[0] = (len-1)/4+1;
        int i,j,k;
        for(i = 1; i < maxn; i++) d[i] = 0;
        for(i = len-1;i>=0;i--){
            j = (len-1-i)/4+1;
            k = (len-1-i)%4;
            d[j]+=ten[k]*(s[i]-'0');
        }
        while(d[0]>1&&d[d[0]]==0) --d[0];
    }
    BigNum(){
        *this = BigNum(string("0"));
    }
    string toString(){
    	string s("");
    	int i,j,tmp;
    	for( i = 3; i >= 1; i--) if(d[d[0]]>=ten[i]) break;
    	tmp = d[d[0]];
    	for(j = i; j>=0; j--){
    		s=s+(char)(tmp/ten[j]+'0');
    		tmp%=ten[j];
    	}
    	for(i = d[0]-1; i>0; i--){
    		tmp = d[i];
    		for(j = 3; j >=0 ;j--){
    			s = s+(char)(tmp/ten[j]+'0');
    			tmp%=ten[j];
    		}
    	}
    	return s;
    }
};
BigNum operator + (BigNum a,BigNum b){
    BigNum c;
    c.d[0] = max(a.d[0],b.d[0]);
    int x=0;
    for(int i = 1; i <= c.d[0]; i++){
        x = a.d[i]+b.d[i]+x;
        c.d[i] = x%10000;
        x/=10000;
    }
    while(x!=0){
        c.d[++c.d[0]]=x%10000;
        x/=10000;
    }
    return c;
}
BigNum operator * (BigNum a,BigNum b){
    BigNum c;
    c.d[0] = a.d[0]+b.d[0];
    for(int i = 1; i<= a.d[0]; i++){
        int x = 0;
        for(int j = 1; j <= b.d[0]; j++){
            x = a.d[i]*b.d[j]+x+c.d[i+j-1];
            c.d[i+j-1]=x%10000;
            x/=10000;
        }
        c.d[i+b.d[0]]=x;
    }
    while((c.d[0]>1)&&(c.d[c.d[0]]==0)) --c.d[0];
    return c;
}
BigNum dp[60][1010];
int n,s;

void init(){
    for(int i = 0; i <= n; i++)
    	for(int j = 0; j<= s; j++){
    		dp[i][j] = BigNum("0");
    	}
}
int main(){
    while(~scanf("%d%d",&n,&s)){
        if(s&1){
            cout<<0<<endl;
            continue;
        }
        else if(s>18*n){
            cout<<0<<endl;
            continue;
        }
        init();
        s/=2;
        for(int i = 0; i <= s; i++){
        	if(i>9)
        		dp[1][i] = BigNum("0");
        	else
        		dp[1][i] = BigNum("1");
        }
        for(int i = 0; i <= n; i++){
        	dp[i][0] = BigNum("1");
        	dp[0][i] = BigNum("0");
        }
        for(int i = 1; i <= n; i++){
        	for(int j = 1; j <= s; j++){
        		for(int k = 0; k <=9; k++){
        			if(j>=k){
        				dp[i][j] = dp[i-1][j-k]+dp[i][j]; 
        			}
        			
        		}
        	}
        }
        BigNum ans = dp[n][s];
        ans = ans*ans;
    	cout<<ans.toString()<<endl;
    }
    return 0;
}