poj1003预处理+二分查找

Hangover

Time Limit: 1000 MS Memory Limit: 10000 KB64-bit integer IO format: %I64d , %I64u Java class name: Main[Submit] [Status] [Discuss]

Description

How far can you make a stack of cards overhang a table? If you have one card, you can create a maximum overhang of half a card length. (We're assuming that the cards must be perpendicular to the table.) With two cards you can make the top card overhang the bottom one by half a card length, and the bottom one overhang the table by a third of a card length, for a total maximum overhang of 1/2 + 1/3 = 5/6 card lengths. In general you can make n cards overhang by 1/2 + 1/3 + 1/4 + ... + 1/(n + 1) card lengths, where the top card overhangs the second by 1/2, the second overhangs tha third by 1/3, the third overhangs the fourth by 1/4, etc., and the bottom card overhangs the table by 1/(n + 1). This is illustrated in the figure below.

Input

The input consists of one or more test cases, followed by a line containing the number 0.00 that signals the end of the input. Each test case is a single line containing a positive floating-point number c whose value is at least 0.01 and at most 5.20; c will contain exactly three digits.

Output

For each test case, output the minimum number of cards necessary to achieve an overhang of at least c card lengths. Use the exact output format shown in the examples.

Sample Input

1.00
3.71
0.04
5.19
0.00

Sample Output

3 card(s)
61 card(s)
1 card(s)
273 card(s)

由于已经知道答案不可能大于5.20，因此预处理len[i]=len[i-1]+1.0/double(i+1)到5.20，再二分查找即可，注意精度

#include<iostream>
#include<stdio.h>
#include <string.h>
#define delta 1e-8
using namespace std;
int ans=0;
double len[1000];
double c;
int zero(double x){
if (x>delta)return 1;
else  if (x<delta)return -1;
else  if (x==delta)return 0;
}
int jug(int l,int r){
if (l+1>r)return r;
int mid=(l+r)/2;
if ( zero(len[mid]-c) > 0){
jug(l,mid);
}else{
jug(mid+1,r);
}
}
int main (){
double lens,sum;
int n=0;
len[0]=0;
for (int i=1;zero(len[i-1]-5.20)<0;i++){
len[i]=len[i-1]+1.0/double(i+1);
n++;
}
int l=1,r=n,mid=(r+l)/2;
while (~scanf("%lf",&c)){
if (c==0.00)break;
ans=jug(l,r);
printf("%d card(s)\n",ans);
}

return 0;
}

Hangover

Time Limit: 1000 MS Memory Limit: 10000 KB64-bit integer IO format: %I64d , %I64u Java class name: Main[Submit] [Status] [Discuss]

Description

How far can you make a stack of cards overhang a table? If you have one card, you can create a maximum overhang of half a card length. (We're assuming that the cards must be perpendicular to the table.) With two cards you can make the top card overhang the bottom one by half a card length, and the bottom one overhang the table by a third of a card length, for a total maximum overhang of 1/2 + 1/3 = 5/6 card lengths. In general you can make n cards overhang by 1/2 + 1/3 + 1/4 + ... + 1/(n + 1) card lengths, where the top card overhangs the second by 1/2, the second overhangs tha third by 1/3, the third overhangs the fourth by 1/4, etc., and the bottom card overhangs the table by 1/(n + 1). This is illustrated in the figure below.

Input

The input consists of one or more test cases, followed by a line containing the number 0.00 that signals the end of the input. Each test case is a single line containing a positive floating-point number c whose value is at least 0.01 and at most 5.20; c will contain exactly three digits.

Output

For each test case, output the minimum number of cards necessary to achieve an overhang of at least c card lengths. Use the exact output format shown in the examples.

Sample Input

1.00
3.71
0.04
5.19
0.00

Sample Output

3 card(s)
61 card(s)
1 card(s)
273 card(s)