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Topic 1034: Looking for a Millionaire time limit:1 seconds Memory limit:32 MB Special question: No submitted:5168 Resolution:2075 Title Description: Zhejiang Tongxiang Wu Town A total of n individuals, please find the town of the former m monopoly. Input: The input contains multiple sets of test cases. Each use case first contains 2 integers n (

Http://acm.ocrosoft.com/problem.php?cid=1316pid=8Title DescriptionA town in Hangzhou, Zhejiang has a total of n individuals, please find the town's former m monopoly.InputThe input contains multiple sets of test cases.Each use case first contains 2 integers n (0N and m at 0 indicate the end of the input.OutputPlease output Wuzhen former M monopoly property number, property of the row front, if the monopoly less than m, then all output, each group of output accounted for a row.Sample input3 11 6

it overrides the smallest digit min in the array, skip, enter the next, and then continue ...After the input and processing is complete, sort in descending order and then output.The code is as follows:1#include 2#include 3 intcmpConst void*a,Const void*b)4 {5 return*(int*) B-* (int*) A;6 }7 intFinminindex (intA[],intLen);//returns the subscript for the smallest element of a array. A the number of array elements is Len8 intMain ()9 {Ten intn,m; One inta[ One]={0}; A inti; - i

, if 50Code implementation: The optimization of spatial complexity has been completed by using a rolling array loop#include #include #include #include using namespace Std;int M, X;Double P;Double dp[2][(1 void Solve (){int n = 1 Double *pre = dp[0], *nxt = dp[1];memset (pre, 0, sizeof (double) * (n + 1));Pre[n] = 1.0;//: module n corresponds to the funds >= 1 millionfor (int r = 0; r {for (int i = 0; I {int step = min (i, n-i);//Avoid i + J > NDouble T = 0.0;for (int j = 0; J t = max (t, P * pre

Broken king becomes a millionaire In the outside world, Curt degerman is a poor hacker. The elderly Swede, nicknamed "Broken King", has been walking through the streets and alleys in the northern Swedish city of schelefteo for 30 years, wearing blue jackets and worn-out trousers, you can change money by using a can or bottle. In the eyes of most people, he is an ordinary street tramp. However, after his death, he left his cousin with a heri

Looking for a MillionaireTime limit:2000/1000 MS (java/others) Memory limit:32768/32768 K (java/others)Total submission (s): 4169 Accepted Submission (s): 1714Problem description Zhejiang Tongxiang Wu Town A total of n people, please find out the town of the former m monopoly.Input inputs contain multiple sets of test cases.Each use case first contains 2 integers n (0N and m at 0 indicate the end of the input.Output please export the wealth of the former m monopoly of Wuzhen, the number of asse

Millionaire rank listtime limit: 1000 ms Memory limit: 131072 kbdescription In the wild universe, there lives a race of clever creature. They have high level civilization and control a lot of planets. As we human, some of us are very rich and the public like to know who are the top 10. as a journalist of a famous new paper, you are asked to write a program to get the top 10 list.Input The input contains only one test case. The first line contains one

Title Description: Zhejiang Tongxiang Wu Town A total of n individuals, please find the town of the former m monopoly. Input: The input contains multiple sets of test cases. Each use case first contains 2 integers n (0N and m at 0 indicate the end of the input. Output: please output wuzhen former M monopoly property number, property of the row front, if the monopoly less than m, then all output, eac

Test instructionsYou have x dollars for the M-wheel gambling game. Each round can be held by any part of the money as a bet (bet 0 yuan for this round), the bet can be fractional, not necessarily an integer. Each round wins the probability is P, wins the wager to double, loses the wager to be gone. If you finally hold at least 1000000 yuan, you can take all the money. It is required to calculate the probability of obtaining at least 1000000 yuan in the optimal strategy.Data range:0111 intM, X; 2

(GCC):#include inta[ -];intm;voidSwap (intXintY//Exchange Order{ intT; T=A[x]; A[X]=A[y]; A[y]=t;}voidSiftdown (intI//Look down{ intt,flag=0;//T is used to record smaller nodes and flag to determine if they can be adjusted downwards . while(i*20) { if(a[i]>a[i*2]) T=i*2; ElseT=i; if(i*2+1m) {if(a[t]>a[i*2+1]) T=i*2+1; } if(t!=i) {Swap (t,i);//Exchangei=t;//Update down } ElseFlag=1;//Otherwise it can't be adjusted. }}voidHeapsort ()//Heap Sort{ whi

/** Copyright (c) 2012, Computer College, Yantai University* All rights reserved.* Author: Tian Yu* Date of Completion: November 4, 2012* Version Number: v1.0** Input Description: None* Problem Description: None* Program output: None* Problem

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