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http://acm.hdu.edu.cn/showproblem.php?pid=5389
Problem DescriptionZero Escape, is a visual novel adventure video game directed by Kotaro Uchikoshi (you may hear about ever17?) and developed by Chunsoft. Stilwell is enjoying the first chapter of this series, and in this chapter digital root is an important factor. This is the definition of digital root on Wikipedia: The digital root of a non-negative integer is the single digit value obtained by an iterative process of summing digits, on each iteration using the result from the previous iteration to compute a digit sum. The process continues until a single-digit number is reached. For example, the digital root of65536 is7 , because6+5+5+3+6=25 and2+5=7 . In the game, every player has a special identifier. Maybe two players have the same identifier, but they are different players. If a group of players want to get into a door numberedX(1≤X≤9) , the digital root of their identifier sum must beX . For example, players{1,2,6} can get into the door9 , but players{2,3,3} can‘t. There is two doors, numberedA andB . MaybeA=B , but they are two different door. And there isn players, everyone must get into one of these two doors. Some players will get into the doorA , and others will get into the doorB . For example: players are{1,2,6} ,A=9 ,B=1 There is only one way to distribute the players: all players get into the door9 . Because there is no player to get into the door1 , the digital root limit of this door will be ignored. Given the identifier of every player, please calculate how many kinds of methods are there,mod 258280327 .InputThe first line of the input contains a single numberT , the number of test cases. For each test case, the first line contains three integersn ,A andB . Next line containsn integersidi , describing the identifier of every player.T≤100 ,n≤105 ,∑n≤106 ,1≤A,B,idi≤9 OutputFor each test case, output a single integer in a single line, the number of ways that thesen players can get into these two doors.Sample Input4 3 9 1 1 2 6 3 9 1 2 3 3 5 2 3 1 1 1 1 1 9 9 9 1 2 3 4 5 6 7 8 9Sample Output1 0 10 60
/**
hdu 5389 dp类似背包
题目大意：给出一组数让分成两组，组成两数的数根分别为a和b，那么有多少种分法
解题思路：这题主要是透彻理解数根的求法，我们一个数的数根为各个位上数的和相加模9如果为0则为9,并且两数和的数根等于两数数根和的数根。
有了这个规律我们可以用类似背包的方法求出。dp[i][j]表示前i个数的数根为j有多少种取法。最后若a和b和的数根为所有数的数根，那么dp[n][a]
便为答案。此外还要考虑全部放于a和全部放于b的结果。状态转移方程见代码
*/
#include <stdio.h>
#include <string.h>
#include <algorithm>
#include <iostream>
using namespace std;
const int maxn=100500;
const int mod=258280327;
int n,a,b,id[maxn];
int dp[maxn][15];
int get_sum(int x,int y)
{
int tmp=(x+y)%9;
if(tmp==0)tmp=9;
return tmp;
}
int main()
{
int T;
scanf("%d",&T);
while(T--)
{
scanf("%d%d%d",&n,&a,&b);
int sum=0;
for(int i=1;i<=n;i++)
{
scanf("%d",&id[i]);
sum=get_sum(sum,id[i]);
}
memset(dp,0,sizeof(dp));
dp[0][0]=1;
for(int i=1;i<=n;i++)
{
for(int j=0;j<=9;j++)
{
dp[i][j]=(dp[i][j]+dp[i-1][j])%mod;
int ans=get_sum(j,id[i]);
dp[i][ans]=(dp[i][ans]+dp[i-1][j])%mod;
}
}
int ans=0;
if(get_sum(a,b)==sum)
{
ans+=dp[n][a];
if(a==sum)ans--;
}
if(a==sum)ans++;
if(b==sum)ans++;
printf("%d\n",ans%mod);
}
return 0;
}
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原文地址：http://blog.csdn.net/lvshubao1314/article/details/47664845
