Codeforeces Round #226 (Div. 2) E---Bear in the Field(矩阵快速幂)
2017-09-06 16:38
417 查看
Our bear's forest has a checkered field. The checkered field is an n × n table, the rows are numbered from 1 to n from
top to bottom, the columns are numbered from 1 ton from left to right. Let's denote a cell of the field on the intersection of row xand column y by
record (x, y). Each cell of the field contains growing raspberry, at that, the cell (x, y) of the field contains x + y raspberry
bushes.
The bear came out to walk across the field. At the beginning of the walk his speed is (dx, dy). Then the bear spends exactly t seconds
on the field. Each second the following takes place:
Let's suppose that at the current moment the bear is in cell (x, y).
First the bear eats the raspberry from all the bushes he has in the current cell. After the bear eats the raspberry from k bushes, he increases each component of his speed by k.
In other words, if before eating the k bushes of raspberry his speed was (dx, dy), then after eating the berry his speed equals(dx + k, dy + k).
Let's denote the current speed of the bear (dx, dy) (it was increased after the previous step). Then the bear moves from cell (x, y) to
cell (((x + dx - 1) mod n) + 1, ((y + dy - 1) mod n) + 1).
Then one additional raspberry bush grows in each cell of the field.
You task is to predict the bear's actions. Find the cell he ends up in if he starts from cell (sx, sy). Assume that each bush has infinitely much raspberry and the bear will never eat all of it.
Input
The first line of the input contains six space-separated integers: n, sx, sy, dx,dy, t (1 ≤ n ≤ 109; 1 ≤ sx, sy ≤ n; - 100 ≤ dx, dy ≤ 100; 0 ≤ t ≤ 1018).
Output
Print two integers — the coordinates of the cell the bear will end up in after tseconds.
Example
Input
Output
Input
Output
Note
Operation a mod b means taking the remainder after dividing a by b. Note
that the result of the operation is always non-negative. For example, ( - 1) mod 3 = 2.
In the first sample before the first move the speed vector will equal (3,4) and the bear will get to cell (4,1). Before the second move the speed vector will equal (9,10) and he bear will get to cell (3,1). Don't forget that at the second move, the number
of berry bushes increased by 1.
In the second sample before the first move the speed vector will equal (1,1) and the bear will get to cell (1,1). Before the second move, the speed vector will equal (4,4) and the bear will get to cell (1,1). Don't forget that at the second move, the number
of berry bushes increased by 1.
123
题意:
n*n的 图 熊初始点在(sx,sy) 速度为 (dx,dy) 每过1 s 都有:
1:速度增加 k k 为 x+y ;
2:每个(x,y) 增加 1
3: 熊走的方向 x>>dx +1, y--> dy +1
根据 关系
我们可以列出:
构造矩阵:
#include <iostream>
#include <stdio.h>
#include <algorithm>
#include <cmath>
#include <cstring>
#include <string>
#include <queue>
#include <stack>
#include <stdlib.h>
#include <list>
#include <map>
#include <set>
#include <bitset>
#include <vector>
#define mem(a,b) memset(a,b,sizeof(a))
#define findx(x) lower_bound(b+1,b+1+bn,x)-b
#define FIN freopen("input.txt","r",stdin)
#define FOUT freopen("output.txt","w",stdout)
#define S1(n) scanf("%d",&n)
#define SL1(n) scanf("%I64d",&n)
#define S2(n,m) scanf("%d%d",&n,&m)
#define SL2(n,m) scanf("%I64d%I64d",&n,&m)
#define Pr(n) printf("%d\n",n)
using namespace std;
typedef long long ll;
const double PI=acos(-1);
const int INF=0x3f3f3f3f;
const double esp=1e-6;
const int maxn=1e6+5;
int dir[5][2]={0,1,0,-1,1,0,-1,0};
const int MAXN=6;
const int N=10;
int mod,sx,sy,dx,dy;
ll t;
struct Matrix{
ll arr
;
void init()
{
memset(arr,0,sizeof(arr));
for(int i=0;i<MAXN;i++)
arr[i][i]=1;//初始化
}
void iinit()
{
memset(arr,0,sizeof(arr));
arr[0][0]=arr[1][1]=arr[0][5]=arr[1][5]=arr[2][5]=arr[3][5]=2;
arr[0][1]=arr[0][2]=arr[0][4]=1;
arr[1][0]=arr[1][3]=arr[1][4]=1;
arr[2][0]=arr[2][1]=arr[2][2]=arr[2][4]=1;
arr[3][0]=arr[3][1]=arr[3][3]=arr[3][4]=1;
arr[4][4]=arr[4][5]=1;
arr[5][5]=1;
}
void obj()
{
memset(arr,0,sizeof(arr));
arr[0][0]=sx-1;
arr[1][0]=sy-1;
arr[2][0]=(dx%mod+mod)%mod;
arr[3][0]=(dy%mod+mod)%mod;
arr[5][0]=1;
}
}A;
Matrix mul(Matrix X,Matrix Y)// 矩阵乘法
{
Matrix ans;
for(int i=0;i<MAXN;i++)
for(int j=0;j<MAXN;j++){
ans.arr[i][j]=0;
for(int k=0;k<MAXN;k++){
ans.arr[i][j]+=X.arr[i][k]*Y.arr[k][j];
ans.arr[i][j]%=mod;
}
}
return ans;
}
Matrix Q_pow(Matrix B,ll n)// 矩阵快速幂
{
Matrix ans;
ans.init();
while(n)
{
if(n&1)
ans=mul(ans,B);
n>>=1;
B=mul(B,B);
}
return ans;
}
int main()
{
while(~scanf("%d %d %d %d %d %lld",&mod,&sx,&sy,&dx,&dy,&t))
{
Matrix ans;
ans.iinit();
ans=Q_pow(ans,t);
Matrix FF;
FF.obj();
FF=mul(ans,FF);
printf("%lld %lld\n",FF.arr[0][0]+1,FF.arr[1][0]+1);
}
return 0;
}
top to bottom, the columns are numbered from 1 ton from left to right. Let's denote a cell of the field on the intersection of row xand column y by
record (x, y). Each cell of the field contains growing raspberry, at that, the cell (x, y) of the field contains x + y raspberry
bushes.
The bear came out to walk across the field. At the beginning of the walk his speed is (dx, dy). Then the bear spends exactly t seconds
on the field. Each second the following takes place:
Let's suppose that at the current moment the bear is in cell (x, y).
First the bear eats the raspberry from all the bushes he has in the current cell. After the bear eats the raspberry from k bushes, he increases each component of his speed by k.
In other words, if before eating the k bushes of raspberry his speed was (dx, dy), then after eating the berry his speed equals(dx + k, dy + k).
Let's denote the current speed of the bear (dx, dy) (it was increased after the previous step). Then the bear moves from cell (x, y) to
cell (((x + dx - 1) mod n) + 1, ((y + dy - 1) mod n) + 1).
Then one additional raspberry bush grows in each cell of the field.
You task is to predict the bear's actions. Find the cell he ends up in if he starts from cell (sx, sy). Assume that each bush has infinitely much raspberry and the bear will never eat all of it.
Input
The first line of the input contains six space-separated integers: n, sx, sy, dx,dy, t (1 ≤ n ≤ 109; 1 ≤ sx, sy ≤ n; - 100 ≤ dx, dy ≤ 100; 0 ≤ t ≤ 1018).
Output
Print two integers — the coordinates of the cell the bear will end up in after tseconds.
Example
Input
5 1 2 0 1 2
Output
3 1
Input
1 1 1 -1 -1 2
Output
1 1
Note
Operation a mod b means taking the remainder after dividing a by b. Note
that the result of the operation is always non-negative. For example, ( - 1) mod 3 = 2.
In the first sample before the first move the speed vector will equal (3,4) and the bear will get to cell (4,1). Before the second move the speed vector will equal (9,10) and he bear will get to cell (3,1). Don't forget that at the second move, the number
of berry bushes increased by 1.
In the second sample before the first move the speed vector will equal (1,1) and the bear will get to cell (1,1). Before the second move, the speed vector will equal (4,4) and the bear will get to cell (1,1). Don't forget that at the second move, the number
of berry bushes increased by 1.
123
题意:
n*n的 图 熊初始点在(sx,sy) 速度为 (dx,dy) 每过1 s 都有:
1:速度增加 k k 为 x+y ;
2:每个(x,y) 增加 1
3: 熊走的方向 x>>dx +1, y--> dy +1
根据 关系
我们可以列出:
构造矩阵:
#include <iostream>
#include <stdio.h>
#include <algorithm>
#include <cmath>
#include <cstring>
#include <string>
#include <queue>
#include <stack>
#include <stdlib.h>
#include <list>
#include <map>
#include <set>
#include <bitset>
#include <vector>
#define mem(a,b) memset(a,b,sizeof(a))
#define findx(x) lower_bound(b+1,b+1+bn,x)-b
#define FIN freopen("input.txt","r",stdin)
#define FOUT freopen("output.txt","w",stdout)
#define S1(n) scanf("%d",&n)
#define SL1(n) scanf("%I64d",&n)
#define S2(n,m) scanf("%d%d",&n,&m)
#define SL2(n,m) scanf("%I64d%I64d",&n,&m)
#define Pr(n) printf("%d\n",n)
using namespace std;
typedef long long ll;
const double PI=acos(-1);
const int INF=0x3f3f3f3f;
const double esp=1e-6;
const int maxn=1e6+5;
int dir[5][2]={0,1,0,-1,1,0,-1,0};
const int MAXN=6;
const int N=10;
int mod,sx,sy,dx,dy;
ll t;
struct Matrix{
ll arr
;
void init()
{
memset(arr,0,sizeof(arr));
for(int i=0;i<MAXN;i++)
arr[i][i]=1;//初始化
}
void iinit()
{
memset(arr,0,sizeof(arr));
arr[0][0]=arr[1][1]=arr[0][5]=arr[1][5]=arr[2][5]=arr[3][5]=2;
arr[0][1]=arr[0][2]=arr[0][4]=1;
arr[1][0]=arr[1][3]=arr[1][4]=1;
arr[2][0]=arr[2][1]=arr[2][2]=arr[2][4]=1;
arr[3][0]=arr[3][1]=arr[3][3]=arr[3][4]=1;
arr[4][4]=arr[4][5]=1;
arr[5][5]=1;
}
void obj()
{
memset(arr,0,sizeof(arr));
arr[0][0]=sx-1;
arr[1][0]=sy-1;
arr[2][0]=(dx%mod+mod)%mod;
arr[3][0]=(dy%mod+mod)%mod;
arr[5][0]=1;
}
}A;
Matrix mul(Matrix X,Matrix Y)// 矩阵乘法
{
Matrix ans;
for(int i=0;i<MAXN;i++)
for(int j=0;j<MAXN;j++){
ans.arr[i][j]=0;
for(int k=0;k<MAXN;k++){
ans.arr[i][j]+=X.arr[i][k]*Y.arr[k][j];
ans.arr[i][j]%=mod;
}
}
return ans;
}
Matrix Q_pow(Matrix B,ll n)// 矩阵快速幂
{
Matrix ans;
ans.init();
while(n)
{
if(n&1)
ans=mul(ans,B);
n>>=1;
B=mul(B,B);
}
return ans;
}
int main()
{
while(~scanf("%d %d %d %d %d %lld",&mod,&sx,&sy,&dx,&dy,&t))
{
Matrix ans;
ans.iinit();
ans=Q_pow(ans,t);
Matrix FF;
FF.obj();
FF=mul(ans,FF);
printf("%lld %lld\n",FF.arr[0][0]+1,FF.arr[1][0]+1);
}
return 0;
}
相关文章推荐
- Codeforces - 385E. 4000 Bear in the Field - 矩阵快速幂
- Codeforces Round #226 (Div. 2) E---Bear in the Field(矩阵)
- 解题报告:Codeforces Round #226 (Div. 2)E. Bear in the Field 矩阵加速幂
- CodeForces 385 E.Bear in the Field(dp+矩阵快速幂)
- Codeforces Round #373 (Div. 2) E. Sasha and Array 矩阵快速幂+线段树
- Codeforces Round #226 (Div. 2) E(矩阵快速幂)
- Codeforces Round #420 (Div. 2) E. Okabe and El Psy Kongroo(矩阵快速幂)
- Codeforces Round #341 (Div. 2)E(矩阵快速幂优化dp,好题)
- POJ 2388 Who's in the Middle (快速选择算法:O(N)求数列第K大)
- HDU 2807 The Shortest Path【矩阵的快速比较】
- Use the Field in Word 2003.
- Codeforces Round #326 Div.1 C.Duff in the Army 树上倍增
- From a response document, how do you update a field in the parent document?
- Codeforces #284 div1 D. Traffic Jams in the Land 数论 线段树
- important, influential, topical or otherwise interesting paper in the field of CS every weekday
- John Paul Mueller, «Windows Administration at the Command Line for Windows 2003, Windows XP, and Windows 2000: In the Field Resu
- HDU 5318 The Goddess Of The Moon(矩阵快速幂)
- 【矩阵快速幂】ZOJ 2974 Just Pour the Water
- 2017 Wuhan University Programming Contest (Online Round) E. Lost in WHU(矩阵快速幂)
- POJ 2388 Who's in the Middle (快速选择算法:O(N)求数列第K大)