Codeforces Round #341 Div.2 C. Wet Shark and Flowers

题意: 不概括了..太长了..

额第一次做这种问题 算是概率dp吗?

保存前缀项中第一个和最后一个的概率 然后每添加新的一项 就解除前缀和第一项和最后一项的关系 并添加新的一项和保存的两项的关系

这里关系指的是两者相邻会产生的额外收入(其中一个满足条件就能得到 因此公式是 2000 * (rate[a] * rate[b] + rate[a] * ( 1 - rate[b]) + rate[b] * (1 - rate[a]))

至于一开始为什么老是调不过去呢..我发现添加第三项的时候前两项是不用解除关系的...

啊啊啊啊这题我敲了大半个小时敲得都烦死了= = 不过交上去直接Pretest Pass了

代码稍显冗长.. 凑和着看吧..

#include <cstdio>
#include <cmath>
#include <cstring>
#include <queue>
#include <vector>
#include <algorithm>
#define INF 0x3f3f3f3f
#define mem(str,x) memset(str,(x),sizeof(str))  
#define STOP puts("Pause");
using namespace std;
typedef long long LL;

int main()
{
    int n, p;
    int l, r;
    double rate[3], ans = 0;
    scanf("%d%d", &n, &p);
    scanf("%d%d", &l, &r);
    if(l % p) rate[0] = (double)(r / p - l / p) / (r - l + 1);
    else rate[0] = (double)(r / p - l / p + 1) / (r - l + 1);
    //printf("i = %d rate = %f
", 1, rate[0]);
    for(int i = 2; i <= n; i++){
        scanf("%d%d", &l, &r);
        if(l % p) rate[2] = (double)(r / p - l / p) / (r - l + 1);
           else rate[2] = (double)(r / p - l / p + 1) / (r - l + 1);
        if(i == 2){
            ans += (rate[0] * rate[2] + rate[0] * (1 - rate[2]) + rate[2] * (1 - rate[0])) * 2000;
            rate[1] = rate[2];
        }
        else if (i == 3){
            ans += (rate[0] * rate[2] + rate[0] * (1 - rate[2]) + rate[2] * (1 - rate[0])) * 2000
                +  (rate[1] * rate[2] + rate[1] * (1 - rate[2]) + rate[2] * (1 - rate[1])) * 2000;
               rate[1] = rate[2];
        }
        else{
            ans += (rate[0] * rate[2] + rate[0] * (1 - rate[2]) + rate[2] * (1 - rate[0])) * 2000
                +  (rate[1] * rate[2] + rate[1] * (1 - rate[2]) + rate[2] * (1 - rate[1])) * 2000
                -  (rate[0] * rate[1] + rate[0] * (1 - rate[1]) + rate[1] * (1 - rate[0])) * 2000;
               rate[1] = rate[2];
        }
        //printf("i = %d rate = %f ans = %f
", i, rate[2], ans);
    }
    printf("%f
", ans);
    return 0;
}