Codeforces Round #437 (Div. 1) C
来源:互联网 发布:软件结项文档 编辑:程序博客网 时间:2024/05/19 22:28
传送门
题意:一个游戏一共有n个关卡,对于第i关,用a[i]时间通过的概率为p[i],用b[i]通过的时间为1-p[i],每通过一关后可以选择继续下一关或者时间清0并从第一关开始,先要求通过所有关卡的时间和不能超过R才算彻底通关,问直到彻底通关位置的游戏时间的期望值为多少
做法:
答案一定是满足递增性质的,那么我们二分答案肯定没毛病呀,另外,这个是个dp,因为他每一关的期望对下一关有影响,所以用dp处理没毛病啊,另外转移方程是什么呢。
动动脑子想一想,如果dp[i][j]代表第i关最后的结果j的期望。
嗨呀,那么dp[i][j]=(dp[i+1][j+f[i]]+f[i])*p[i]/100+(dp[i+1][i+s[i]]+s[i])*(100-p[i])/100了嘛,最后dp[0][0]就为答案了呀。
/// .-~~~~~~~~~-._ _.-~~~~~~~~~-./// __.' ~. .~ `.__/// .'// \./ \\`./// .'// | \\`./// .'// .-~"""""""~~~~-._ | _,-~~~~"""""""~-. \\`./// .'//.-" `-. | .-' "-.\\`./// .'//______.============-.. \ | / ..-============.______\\`./// .'______________________________\|/______________________________`.#pragma comment(linker, "/STACK:1024000000,1024000000")#include <vector>#include <iostream>#include <string>#include <map>#include <stack>#include <cstring>#include <queue>#include <list>#include <stdio.h>#include <set>#include <algorithm>#include <cstdlib>#include <cmath>#include <iomanip>#include <cctype>#include <sstream>#include <functional>#include <stdlib.h>#include <time.h>#include <bitset>using namespace std;#define pi acos(-1)#define s_1(x) scanf("%d",&x)#define s_2(x,y) scanf("%d%d",&x,&y)#define s_3(x,y,z) scanf("%d%d%d",&x,&y,&z)#define s_4(x,y,z,X) scanf("%d%d%d%d",&x,&y,&z,&X)#define S_1(x) scan_d(x)#define S_2(x,y) scan_d(x),scan_d(y)#define S_3(x,y,z) scan_d(x),scan_d(y),scan_d(z)#define PI acos(-1)#define endl '\n'#define srand() srand(time(0));#define me(x,y) memset(x,y,sizeof(x));#define foreach(it,a) for(__typeof((a).begin()) it=(a).begin();it!=(a).end();it++)#define close() ios::sync_with_stdio(0); cin.tie(0);#define FOR(x,n,i) for(int i=x;i<=n;i++)#define FOr(x,n,i) for(int i=x;i<n;i++)#define fOR(n,x,i) for(int i=n;i>=x;i--)#define fOr(n,x,i) for(int i=n;i>x;i--)#define W while#define sgn(x) ((x) < 0 ? -1 : (x) > 0)#define bug printf("***********\n");#define db double#define ll long long#define mp make_pair#define pb push_backtypedef long long LL;typedef pair <int, int> ii;const int INF=0x3f3f3f3f;const LL LINF=0x3f3f3f3f3f3f3f3fLL;const int dx[]={-1,0,1,0,1,-1,-1,1};const int dy[]={0,1,0,-1,-1,1,-1,1};const int maxn=5e+500;const int maxx=4e5+10;const double EPS=1e-8;const double eps=1e-8;const int mod=1e9+7;template<class T>inline T min(T a,T b,T c) { return min(min(a,b),c);}template<class T>inline T max(T a,T b,T c) { return max(max(a,b),c);}template<class T>inline T min(T a,T b,T c,T d) { return min(min(a,b),min(c,d));}template<class T>inline T max(T a,T b,T c,T d) { return max(max(a,b),max(c,d));}template <class T>inline bool scan_d(T &ret){char c;int sgn;if (c = getchar(), c == EOF){return 0;}while (c != '-' && (c < '0' || c > '9')){c = getchar();}sgn = (c == '-') ? -1 : 1;ret = (c == '-') ? 0 : (c - '0');while (c = getchar(), c >= '0' && c <= '9'){ret = ret * 10 + (c - '0');}ret *= sgn;return 1;}inline bool scan_lf(double &num){char in;double Dec=0.1;bool IsN=false,IsD=false;in=getchar();if(in==EOF) return false;while(in!='-'&&in!='.'&&(in<'0'||in>'9'))in=getchar();if(in=='-'){IsN=true;num=0;}else if(in=='.'){IsD=true;num=0;}else num=in-'0';if(!IsD){while(in=getchar(),in>='0'&&in<='9'){num*=10;num+=in-'0';}}if(in!='.'){if(IsN) num=-num;return true;}else{while(in=getchar(),in>='0'&&in<='9'){num+=Dec*(in-'0');Dec*=0.1;}}if(IsN) num=-num;return true;}void Out(LL a){if(a < 0) { putchar('-'); a = -a; }if(a >= 10) Out(a / 10);putchar(a % 10 + '0');}void print(LL a){ Out(a),puts("");}//freopen( "in.txt" , "r" , stdin );//freopen( "data.txt" , "w" , stdout );//cerr << "run time is " << clock() << endl;int n,r;int f[55],s[55],p[55];db dp[55][maxn];int check(db x){ for(int i=n-1;i>=0;i--) { FOr(r+1,maxn,j) dp[i+1][j]=x; FOR(0,r,j) { double t1=(dp[i+1][j+f[i]]+f[i])*p[i]/100; double t2=(dp[i+1][j+s[i]]+s[i])*(100-p[i])/100; dp[i][j]= min(x,t1 + t2); } } return dp[0][0]<x;}void solve(){ W(scanf("%d%d",&n,&r)!=EOF) { FOr(0,n,i) s_3(f[i],s[i],p[i]); db l=0,rr=1e10; int len=100; W(len--) { db mid=(l+rr)/2; if(check(mid)) rr=mid; else l=mid; } printf("%.10f\n",l); }}int main(){ //freopen( "in.txt" , "r" , stdin ); //freopen( "data.txt" , "w" , stdout ); int t=1; //s_1(t); for(int cas=1;cas<=t;cas++) { //printf("Case %d:\n",cas); solve(); }}
阅读全文
0 0
- Codeforces Round #437 (Div. 1) C
- 【dp】Codeforces Round #110 (Div. 1) C
- Codeforces Round #142 (Div. 1) C. Triangles
- Codeforces Round #230 (Div. 1)B,C
- Codeforces Round #259 (Div. 1)C题
- Codeforces Round #228 (Div. 1) C 贪心
- Codeforces Round #284 (Div. 1) C
- Codeforces Round #286 (Div. 1) C、D
- Codeforces Round #302 (Div. 1) C
- Codeforces Round #333 (Div. 1) C
- Codeforces Round #336 (Div. 1) C. Marbles
- Codeforces Round #356 (Div. 1) C
- Codeforces Round #362 (Div. 1) C PLEASE
- Codeforces Round #363 (Div. 1) C LRU
- Codeforces Round #363 (Div. 1) C LRU
- Codeforces Round #364 (Div. 1) C
- Codeforces Round #372 (Div. 1) C
- Codeforces Round #381 (Div. 1) C
- JavaScript字符串操作
- AndroidStudio 设置全局查找快捷键
- js对象深拷贝
- Linux中对进程的操作
- vue 插件集合
- Codeforces Round #437 (Div. 1) C
- 【LeetCode】C# 36、Valid Sudoku
- css的overflow属性
- Util-BaseApplication
- 【第五周项目4】数制转换
- PPT制作流程 --- 早看早超生
- 黑客工具软件大全100套
- 合并有序单链表并排序(遍历一次)
- Studio像Eclipse一样设置switch/case那样的代码块