## Thursday, July 28, 2016

### Not So Random - Round-E APAC Test of Google 2016 (Problem C)

http://www.dss886.com/algorithm/2016/05/18/21-36
There is a certain “random number generator” (RNG) which takes one nonnegative integer as input and generates another nonnegative integer as output. But you know that the RNG is really not very random at all! It uses a fixed number K, and always performs one of the following three operations:
• with probability A/100: return the bitwise AND of the input and K
• with probability B/100: return the bitwise OR of the input and K
• with probability C/100: return the bitwise XOR of the input and K (You may assume that the RNG is truly random in the way that it chooses the operation each time, based on the values of A, B, and C.)
You have N copies of this RNG, and you have arranged them in series such that output from one machine will be the input for the next machine in the series. If > you provide X as an input to the first machine, what will be the expected value of the output of the final machine in the series?
Input
The first line of the input gives the number of test cases, T. T test cases follow; each consists of one line with six integers N, X, K, A, B, and C. Respectively, these denote the number of machines, the initial input, the fixed number with which all the bitwise operations will be performed (on every machine), and 100 times the probabilities of the bitwise AND, OR, and XOR operations.
Output
For each test case, output one line containing “Case #x: y”, where x is the test case number (starting from 1) and y is the expected value of the final output. y will be considered correct if it is within an absolute or relative error of 10^-9 of the correct answer. See the FAQ for an explanation of what that means, and what formats of real numbers we accept.
Limits
1 ≤ T ≤ 50.
0 ≤ A ≤ 100.
0 ≤ B ≤ 100.
0 ≤ C ≤ 100.
A+B+C = 100.
Small dataset
1 ≤ N ≤ 10.
0 ≤ X ≤ 10^4.
0 ≤ K ≤ 10^4.
Large dataset
1 ≤ N ≤ 10^5.
0 ≤ X ≤ 10^9.
0 ≤ K ≤ 10^9.
Sample
InputOutput
3
1 5 5 10 50 40Case #1: 3.0000000000
2 5 5 10 50 40Case #2: 3.6000000000
10 15 21 70 20 10Case #3: 15.6850579098
In sample test case #1, the final output will be 5 if AND or OR happens and 0 if XOR happens. So the probability of getting 5 is (0.1 + 0.5) and the probability of getting 0 is 0.4. So the expected final output is 5 * 0.6 + 0 * 0.4 = 3.
In sample test case #2, the final output will be 5 with probability 0.72, and 0 otherwise.

### 单个RNG

ABC
0 & 0 = 0　　0 | 0 = 0　　0 ^ 0 = 0
0 & 1 = 0　　0 | 0 = 1　　0 ^ 1 = 1
1 & 0 = 0　　0 | 0 = 1　　1 ^ 0 = 1
1 & 1 = 1　　1 | 1 = 1　　1 ^ 1 = 0

k = 0:
[　100 ,　 0　]
[　A　,　B+C　]
k = 1:
[　A　,　B + C　]
[　C　,　A + B　]
（即k=1，输入为0时，输出A%为0、(B+C)%为1；输入为1时，输出C%为0，(A+B)%为1）

### 多个RNG串联

* From the Round-E APAC Test of Google 2016 (Problem C)
*
* The main solving ideas:
* 1. All Three operation is bit-based, so we can look into the operation of single bit.
* 2. Instead of considering the 3 operations and their probabilities, we can just think about the input and output
*    probability of 0-1, then we got two matrices as below:
*        k = 0: [ 100 ,  0   ] , k = 1: [  A  , B + C ]
*               [  A  , B+C  ]          [  C  , A + B ]
*    (which means: when x=0 and k=1, the output got 0 by chance of A%, and 1 by chance of (B+C)%)
* 3. When have N copies of this RNG in series, we multiply the matrices to itself by N-1 times, then we got the total
*    system's output matrices.
* 4. Now for every bit of X and K, we can calculate the output bit chance from the matrices above, then we can finally
*    got the result.
*
* For example:
*    N=1, X=5, K=5, A=10, B=50, C=40, the single and total matrices (N=1) is:
*        k = 0: [ 100 ,  0  ] , k = 1: [  10  , 90 ]
*               [ 10  , 90  ]          [  40  , 60 ]
*    X = 00000000000000000000000000000101
*    K = 00000000000000000000000000000101
*    1. when bit of X and K are both 0, the output bit is 0 by 100%.
*    2. when bit of X and K are both 1, the output bit is 0 by 40% and 1 by 60%.
*    So the expect value of result is (60% * 4 + 60% * 1), which is 3.
*
* For Improvement:
*   The main time-consuming part of this solution is the multiplication of matrices.
*   Some fast matrix multiplication like the Strassen-Algorithm will reduce the consumed time.
*/
public class NotSoRandom {
public static void main(String[] args) {
Scanner scanner = new Scanner(System.in);
while (scanner.hasNext()) {
int T = scanner.nextInt();
for (int t = 0; t < T; t++) {
int N = scanner.nextInt();
int X = scanner.nextInt();
int K = scanner.nextInt();
int A = scanner.nextInt();
int B = scanner.nextInt();
int C = scanner.nextInt();
Matrix bitZero = new Matrix(100, 0, A, B + C);
Matrix bitOne = new Matrix(A, B + C, C, A + B);
Matrix bitZeroAfterK = bitZero;
Matrix bitOneAfterK = bitOne;
for (int n = 1; n < N; n++) {   // Maybe some Fast-Matrix-Multiplication will do it faster.
bitZeroAfterK = multiply(bitZeroAfterK, bitZero);
bitOneAfterK = multiply(bitOneAfterK, bitOne);
}
double result = 0;
for (int i = 31; i >= 0; i--) {
result *= 2;
Matrix matrix = isLastZero(K, i) ? bitZeroAfterK : bitOneAfterK;
double chanceOfOne = (isLastZero(X, i) ? matrix.b : matrix.d) / 100;
result += chanceOfOne;
}
print(t, result);
}
}
}

private static boolean isLastZero(int num, int position) {
return (num >>> position) % 2 == 0;
}

private static void print(int t, double result) {
System.out.println("Case #" + (t + 1) + ": " + result);
}

private static Matrix multiply(Matrix m1, Matrix m2) {
double a = (m1.a * m2.a + m1.b * m2.c) / 100;
double b = (m1.a * m2.b + m1.b * m2.d) / 100;
double c = (m1.c * m2.a + m1.d * m2.c) / 100;
double d = (m1.c * m2.b + m1.d * m2.d) / 100;
return new Matrix(a, b, c, d);
}

private static class Matrix {
private double a, b, c, d;
public Matrix(double a, double b, double c, double d) {
this.a = a;
this.b = b;
this.c = c;
this.d = d;
}
}

If (k & (1 << j)) > 0 :
dp[i][j][0] += dp[i-1][j][0] * a / 100
dp[i][j][0] += dp[i-1][j][1] * c / 100
dp[i][j][1] += dp[i-1][j][1] * a / 100
dp[i][j][1] += dp[i-1][j][0] * b / 100
dp[i][j][1] += dp[i-1][j][1] * b / 100
dp[i][j][1] += dp[i-1][j][0] * c / 100
Else :
dp[i][j][0] += dp[i-1][j][0] * a / 100
dp[i][j][0] += dp[i-1][j][1] * a / 100
dp[i][j][0] += dp[i-1][j][0] * b / 100
dp[i][j][0] += dp[i-1][j][0] * c / 100
dp[i][j][1] += dp[i-1][j][1] * b / 100
dp[i][j][1] += dp[i-1][j][1] * c / 100

typedef long long lld;

const int N = 111111;
const int M = 31; // x & k >= 0, bit(31) = 0

double dp[N][M][2];

double solve()
{
    double ret = 0.0;
    int n, x, k, a, b, c;
    cin >> n >> x >> k >> a >> b >> c;

    // init
    clr(dp, 0);
    for (int j = 0; j < M; ++j) {
        if ( x & (1 << j) ) {
            dp[0][j][0] = 0.0;
            dp[0][j][1] = 1.0;
        } else {
            dp[0][j][0] = 1.0;
            dp[0][j][1] = 0.0;
        }
    }

    // dp
    for (int j = 0; j < M; ++j) {
        for (int i = 1; i <= n; ++i) {
            if ( k & (1 << j) ) {
                dp[i][j][0] += dp[i-1][j][0] * a / 100;
                dp[i][j][0] += dp[i-1][j][1] * c / 100;
                dp[i][j][1] += dp[i-1][j][1] * a / 100;
                dp[i][j][1] += (dp[i-1][j][0] + dp[i-1][j][1]) * b / 100;
                dp[i][j][1] += dp[i-1][j][0] * c / 100;
            } else {
                dp[i][j][0] += (dp[i-1][j][0] + dp[i-1][j][1]) * a / 100;
                dp[i][j][0] += dp[i-1][j][0] * b / 100;
                dp[i][j][0] += dp[i-1][j][0] * c / 100;
                dp[i][j][1] += dp[i-1][j][1] * b / 100;
                dp[i][j][1] += dp[i-1][j][1] * c / 100;
            }
        }
        ret += dp[n][j][1] * (1 << j);
    }

    return ret;
}

int main ()
{
    freopen("F:/#test-data/in.txt", "r", stdin);
    freopen("F:/#test-data/out.txt", "w", stdout);
    ios::sync_with_stdio(false); cin.tie(0);
    cout << fixed << showpoint;
    int t; cin >> t;
    for (int cas = 1; cas <= t; ++cas) {
        cout << "Case #" << cas << ": ";
        cout << setprecision(9) << solve() << endl;
    }
    return 0;
}
http://blog.csdn.net/ww32zz/article/details/51362183

1. void multiply(double m1[], double m2[]){
2.     double ans[] = {0, 0, 0, 0};
3.     ans[0] = m1[0] * m2[0] + m1[1] * m2[2];
4.     ans[1] = m1[0] * m2[1] + m1[1] * m2[3];
5.     ans[2] = m1[2] * m2[0] + m1[3] * m2[2];
6.     ans[3] = m1[2] * m2[1] + m1[3] * m2[3];
7.     for(int i = 0; i < 4; ++i)
8.         m1[i] = ans[i];
9. }
10.
11. int main(){
12.     int tc, ca = 0;
13.     cin >> tc;
14.     while(tc--){
15.         int n, x, k;
16.         double a, b, c;
17.         cin >> n >> x >> k >> a >> b >> c;
18.         a /= 100, b /= 100, c /= 100;
19.         /*ans0，ans1保存最终的状态*/
20.         double ans0[] = {1, 0, 0, 1}, ans1[] = {1, 0, 0, 1};
21.         double cur0[] = {1, a, 0, b + c}, cur1[] = {a, c, b + c, a + b};
22.         while(n){
23.             if(n & 1)
24.                 multiply(ans0, cur0), multiply(ans1, cur1);
25.             multiply(cur0, cur0), multiply(cur1, cur1);
26.             n >>= 1;
27.         }
28.         /*按位求期望*/
29.         int base = 1;
30.         double res = 0;
31.         while(x || k){
32.             /*p0，p1表示初始状态为0,1的概率,c0，c1表示相应的矩阵系数*/
33.             double p1 = x & 1 ? 1 : 0, p0 = 1 - p1;
34.             double c0, c1;
35.             if(k & 1)
36.                 c0 = ans1[2], c1 = ans1[3];
37.             else
38.                 c0 = ans0[2], c1 = ans0[3];
39.             res += (c0 * p0 + c1 * p1) * base;  //求解当前bit的期望
40.             base <<= 1, x >>= 1, k >>= 1;
41.         }
42.         printf("Case #%d: %.10lf\n", ++ca, res);
43.     }
44. }