CodeForces 148D Bag of Mice 概率DP

CF的题都喜欢加一堆背景？
锻炼英语翻译233。
第一段不翻了。。。
两人轮流从袋子里（包含w和白老鼠和b个黑老鼠）拿老鼠，第一个拿到白老鼠的人赢。
每次拿走一只老鼠以后，剩下的老鼠就会很恐慌，然后就会有一个老鼠自己跑出来（公主拿老鼠很小心，所以并不会惊动其他的老鼠。公主先拿。公主赢得概率有多大？
如果包里没有更多老鼠，而且没人拿到白老鼠，龙赢。自己跑出去的老鼠不算在拿出来的老鼠里，而且不会回到袋子里。每个老鼠每次拿取都有相等的概率(即1/num)。自己跑出来的概率也相等(即1/num)。

令 fi,j 等到公主拿的时候，剩下j只白老鼠和k只黑老鼠的概率有多大。
包含情况：

• 公主拿走白的，概率为 ii+j 。
• 公主拿走黑的，龙拿走黑的，逃了白的，概率为 ji+j⋅j−1i+j−1⋅ii+j−2⋅fi−1,j−2 。
• 公主拿走黑的，龙拿走黑的，逃了黑的，概率为 ji+j⋅j−1i+j−1⋅j−2i+j−2⋅fi,j−3 。
• 初始状态 fi,0=1,f0,i=0 ，即只有黑老鼠或只有白老鼠的情况。
  好像有人没看懂？逆着递推不会写可以写记忆化搜索的嘛。。

#include <cstdio>
#define FOR(i,j,k) for(i=j;i<=k;++i)
int main() {
    static double dp[1005][1005];
    int w, b, i, j;
    scanf("%d%d", &w, &b);
    FOR(i,1,w) dp[i][0] = 1;
    FOR(i,0,b) dp[0][i] = 0;
    FOR(i,1,w) FOR(j,1,b) {
        dp[i][j] = 1.0 * i / (i + j);
        if (j >= 2) dp[i][j] += 1.0 * j / (i + j) * (j - 1) / (i + j - 1) * i       / (i + j - 2) * dp[i - 1][j - 2];
        if (j >= 3) dp[i][j] += 1.0 * j / (i + j) * (j - 1) / (i + j - 1) * (j - 2) / (i + j - 2) * dp[i    ][j - 3];
    }
    printf("%.9lf", dp[w][b]);
    return 0;
}

D. Bag of mice

time limit per test2 seconds
memory limit per test256 megabytes
inputstandard input
outputstandard output
The dragon and the princess are arguing about what to do on the New Year’s Eve. The dragon suggests flying to the mountains to watch fairies dancing in the moonlight, while the princess thinks they should just go to bed early. They are desperate to come to an amicable agreement, so they decide to leave this up to chance.

They take turns drawing a mouse from a bag which initially contains w white and b black mice. The person who is the first to draw a white mouse wins. After each mouse drawn by the dragon the rest of mice in the bag panic, and one of them jumps out of the bag itself (the princess draws her mice carefully and doesn’t scare other mice). Princess draws first. What is the probability of the princess winning?

If there are no more mice in the bag and nobody has drawn a white mouse, the dragon wins. Mice which jump out of the bag themselves are not considered to be drawn (do not define the winner). Once a mouse has left the bag, it never returns to it. Every mouse is drawn from the bag with the same probability as every other one, and every mouse jumps out of the bag with the same probability as every other one.

Input

The only line of input data contains two integers w and b (0 ≤ w, b ≤ 1000).

Output

Output the probability of the princess winning. The answer is considered to be correct if its absolute or relative error does not exceed 10 - 9.

Examples

input
1 3
output
0.500000000
input
5 5
output
0.658730159

Note

Let’s go through the first sample. The probability of the princess drawing a white mouse on her first turn and winning right away is 1/4. The probability of the dragon drawing a black mouse and not winning on his first turn is 3/4 * 2/3 = 1/2. After this there are two mice left in the bag — one black and one white; one of them jumps out, and the other is drawn by the princess on her second turn. If the princess’ mouse is white, she wins (probability is 1/2 * 1/2 = 1/4), otherwise nobody gets the white mouse, so according to the rule the dragon wins.