Zero Escape
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 600 Accepted Submission(s): 301
Problem Description
Zero 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 of 65536 is 7, because 6+5+5+3+6=25 and 2+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 numbered X(1≤X≤9), the digital root of their identifier sum must be X.
For example, players {1,2,6} can get into the door 9, but players {2,3,3} can’t.
There is two doors, numbered A and B. Maybe A=B, but they are two different door.
And there is n players, everyone must get into one of these two doors. Some players will get into the door A, and others will get into the door B.
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 door 9. Because there is no player to get into the door 1, 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.
Input
The first line of the input contains a single number T, the number of test cases.
For each test case, the first line contains three integers n, A and B.
Next line contains n integers idi, describing the identifier of every player.
T≤100, n≤105, ∑n≤106, 1≤A,B,idi≤9
Output
For each test case, output a single integer in a single line, the number of ways that these n players can get into these two doors.
Sample Input
4
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 9
Sample Output
1
0
10
60
0-1揹包問題，給出n個人的id，有兩個門，每個門有一個標號，我們記作a和b，現在我們要將n個人分成兩組，進入兩個門中，使得兩部分人的標號的和的樹根(sum-1)%9+1分別等於a和b，問有多少種分法。
設dp[i][j]爲當前樹根爲 j 的分類的個數，則狀態轉移就是用當前的值c[i] 和上次的dp[i-1][0]~dp[i-1][9]
dp[i][(j+c[i]-1)%9+1] += dp[i-1][j]; 這是又當下數字之和之前狀態得到的。
dp[i][j] = dp[i-1][j]; 這是通過上次得到的
#include<algorithm>
#include<iostream>
#include<cstring>
#include<cstdio>
#define maxn 100005
#define ll __int64
#define INF 0xffffff
using namespace std;
ll dp[maxn][10]; //第i次
ll c[maxn];
const ll mod=258280327;
int main()
{
int T;
scanf("%d",&T);
while(T--)
{
int n,a,b;
//輸入
scanf("%d%d%d",&n,&a,&b);
for(int i=1; i<=n; i++) scanf("%I64d",&c[i]);
memset(dp,0,sizeof(dp));
dp[0][0]=1;
ll sum=0;
for(int i=1; i<=n; i++)
{
sum+=c[i];
for(int j=0; j<10; j++) dp[i][j]=dp[i-1][j];
for(int j=0; j<10; j++)
{
dp[i][(j+c[i]-1)%9+1]+=dp[i-1][j];
dp[i][(j+c[i]-1)%9+1]%=mod;
}
}
ll ans=0;
if(((sum-1)%9+1)==(a+b-1)%9+1)
{
ans+=dp[n][a];
if((sum-1)%9+1==a)
ans--;
}
if(a==(sum-1)%9+1) ans++;
if(b==(sum-1)%9+1) ans++;
printf("%I64d\n",ans);
}
return 0;
}