hdu5389 Zero Escape(动态规划)
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题目:
Zero Escape
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 131072/131072 K (Java/Others)Total Submission(s): 1209 Accepted Submission(s): 594
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 of65536 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 numberedX(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, numberedA and B . Maybe A=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 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 door9 . 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 .
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
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
For example, players
There is two doors, numbered
And there is
For example:
players are
There is only one way to distribute the players: all players get into the door
Given the identifier of every player, please calculate how many kinds of methods are there,
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 integersn ,A and B .
Next line containsn integers idi , describing the identifier of every player.
T≤100 ,n≤105 ,∑n≤106 ,1≤A,B,idi≤9
For each test case, the first line contains three integers
Next line contains
Output
For 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 Input
43 9 11 2 63 9 12 3 35 2 31 1 1 1 19 9 91 2 3 4 5 6 7 8 9
Sample Output
101060
Author
SXYZ
Source
2015 Multi-University Training Contest 8
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wange2014
题意:有n个人,每个人有自己的编号,可以相同,有两个门也有自己的编号,可以相同,一群人能够通过一个门当且仅当他们的编号之和的数字根等于门的编号,问有多少种使得这n个人通过这两个门的方案。
思路:动态规划,01背包变种,每个人要么选A门,要么选B门,dp[i][j]表示前i个人数字根为j的方案数,那么转移方程就是dp[i][j]=dp[i-1][j]+dp[i-1][(j-idx(i)+9)%9],最终答案就是dp[n][A],注意如果可以全部从B门通过,答案应该加一。
代码:
#include <cstdlib>#include <cctype>#include <cstring>#include <cstdio>#include <cmath>#include<climits>#include <algorithm>#include <vector>#include <string>#include <iostream>#include <sstream>#include <map>#include <set>#include <queue>#include <stack>#include <fstream>#include <numeric>#include <iomanip>#include <bitset>#include <list>#include <stdexcept>#include <functional>#include <utility>#include <ctime>using namespace std;#define PB push_back#define MP make_pair#define REP(i,x,n) for(int i=x;i<(n);++i)#define FOR(i,l,h) for(int i=(l);i<=(h);++i)#define FORD(i,h,l) for(int i=(h);i>=(l);--i)#define SZ(X) ((int)(X).size())#define ALL(X) (X).begin(), (X).end()#define RI(X) scanf("%d", &(X))#define RII(X, Y) scanf("%d%d", &(X), &(Y))#define RIII(X, Y, Z) scanf("%d%d%d", &(X), &(Y), &(Z))#define DRI(X) int (X); scanf("%d", &X)#define DRII(X, Y) int X, Y; scanf("%d%d", &X, &Y)#define DRIII(X, Y, Z) int X, Y, Z; scanf("%d%d%d", &X, &Y, &Z)#define OI(X) printf("%d",X);#define RS(X) scanf("%s", (X))#define MS0(X) memset((X), 0, sizeof((X)))#define MS1(X) memset((X), -1, sizeof((X)))#define LEN(X) strlen(X)#define F first#define S second#define Swap(a, b) (a ^= b, b ^= a, a ^= b)#define Dpoint strcut node{int x,y}#define cmpd int cmp(const int &a,const int &b){return a>b;} /*#ifdef HOME freopen("in.txt","r",stdin); #endif*/const int MOD = 1e9+7;typedef vector<int> VI;typedef vector<string> VS;typedef vector<double> VD;typedef long long LL;typedef pair<int,int> PII;//#define HOMEint Scan(){int res = 0, ch, flag = 0;if((ch = getchar()) == '-')//判断正负flag = 1;else if(ch >= '0' && ch <= '9')//得到完整的数res = ch - '0';while((ch = getchar()) >= '0' && ch <= '9' )res = res * 10 + ch - '0';return flag ? -res : res;}/*----------------PLEASE-----DO-----NOT-----HACK-----ME--------------------*/#define MAXN 100000#define mod 258280327int id[MAXN+5];int dp[MAXN+5][10];int main(){int T;RI(T);while(T--){ int n,A,B; RIII(n,A,B); int sum=0; for(int i=1;i<=n;i++) {RI(id[i]); sum+=id[i];} sum%=9; MS0(dp); dp[0][0]=1; for(int i=1;i<=n;i++) for(int j=0;j<=9;j++) dp[i][j]=(dp[i-1][((j-id[i])%9+9)%9]+dp[i-1][j])%mod; int ans=dp[n][A]; if(sum==B%9) ans=(ans+1)%mod; printf("%d\n",ans);} return 0;}
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