https://www.luogu.org/problemnew/show/CF1096G 显然dp出用\(\frac{n}{2}\)个数能拼出来的每个数的方案数,平方相加就行了,dp显然ntt+快速幂乱搞就做完了 #include<bits/stdc++.h> #define il inline #define vd void #define mod 998244353 typedef long long ll; il int gi(){ int x=0,f=1; char ch=getch…

\(\color{#0066ff}{ 题目描述 }\) 一个\(n\)位数,每位可以是给出的\(k\)个数码中的一个数,可以有前导\(0\),输出前\(n/2\)位之和与后\(n/2\)位之和相等的方案数,保证\(n\)是偶数. \(\color{#0066ff}{输入格式}\) 输入的第一行是两个整数\(n,k\) 接下来的一行有\(k\)个数\(d_1,d_2,\cdots,d_k(0\leq d_i\leq 9)\) \(\color{#0066ff}{输出格式}\) 输出一个数,为方案数…

Lucky ticketsTime Limit: 20 Sec Memory Limit: 256 MB 题目连接 http://acm.hust.edu.cn/vjudge/contest/view.action?cid=86686#problem/J Description Everyone knows that in less than a month the Ice Hockey World Championship will start in Minsk. But few of the…

Lucky tickets Time Limit: 1000MS Memory Limit: 65536K Total Submissions: 3298 Accepted: 2174 Description The public transport administration of Ekaterinburg is anxious about the fact that passengers don't like to pay for passage doing their best to a…

All bus tickets in Berland have their numbers. A number consists of n digits (n is even). Only k decimal digits d1,d2,…,dk can be used to form ticket numbers. If 0 is among these digits, then numbers may have leading zeroes. For example, if n=4 and o…

题目传送门 /* 题意:转换就是求n位数字,总和为s/2的方案数 DP+高精度:状态转移方程:dp[cur^1][k+j] = dp[cur^1][k+j] + dp[cur][k]; 高精度直接拿JayYe的:) 异或运算的规则: 0⊕0=0,0⊕1=1 1⊕0=1,1⊕1=0 口诀:相同取0,相异取1 */ #include <cstdio> #include <cstring> #include <string> #include <iostream>…

Lucky Tickets Time Limit: 2000ms Memory Limit: 16384KB This problem will be judged on Ural. Original ID: 103664-bit integer IO format: %lld      Java class name: (Any)   You are given a number 1 ≤ N ≤ 50. Every ticket has its 2N-digit number. We call…

Lucky tickets Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 3247   Accepted: 2136 Description The public transport administration of Ekaterinburg is anxious about the fact that passengers don't like to pay for passage doing their best…

链接 dp[i][j] += dp[i-1][j-g];背包吧 数据太大了 还是JAVA好用 import java.io.*; import java.math.*; import java.text.*; import java.util.*; public class Big { public static void main(String[] args) { Scanner cin = new Scanner (System.in); BigInteger[][] dp; dp = ][…

题目链接:http://poj.org/problem?id=2346 思路分析:使用动态规划解法:设函数 d( n, x )代表长度为n且满足左边n/2位的和减去右边n/2位的和为x的数的数目. 将一个长度为n的数看做n个数字 A1, A2....An ( 0 <= Ai <= 9  ),将字符分为两个集合{ A1, A2....An/2 } 与 { An/2+1.....An }; 问题转换为求集合中元素和相等的数目.先从每个集合中选出一个元素,求它们差为a的可能数目,即d( 2, a )…