3

PowderHan LV 10 @ 2017-05-07 22:19:01

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstring>
using namespace std;

char rd;    int pn;
template<typename Type>
inline void read(Type& v)
{
    pn=1;
    while((rd=getchar())<'0'||rd>'9')
        if(rd=='-')
            pn=-1;
    v=rd-'0';
    while((rd=getchar())>='0'&&rd<='9')
        v=v*10+rd-'0';
    v*=pn;
}
template<typename Type>
inline void out(Type v,bool c=1)
{
    if(v==0)
        putchar(48);
    else  
    {
        if(v<0)
        {
            putchar('-');
            v=-v;
        }
        int len=0,dg[20];  
        while(v>0)
        {
            dg[++len]=v%10;
            v/=10;
        }  
        for(int i=len;i>=1;i--)
            putchar(dg[i]+48);  
    }
    if(c)
        putchar('\n');
    else
        putchar(' ');
}

double p, q;

int main()
{   
    scanf("%lf%lf", &p, &q);
    p/=100; q/=100;
    for(int i = 0;;i++)
        if(int(q*i-0.000001)-int(p*i+0.000001)>=1)
        {  
            printf("%d\n", i);
            return 0;
        }
    return 0;
}

2

猫粮寸断 LV 10 @ 2019-06-26 22:59:03

//这道题显然使用二分来做，下面说一下二分的合理性
//因为小数只有两位数，所以在p，q不相等时最多有1万人
//p，q若相等，也最多有10万人（实际上到不了）
//因此二分是完全可行的
#include<iostream>
#include<cstdio>
using namespace std;
int main()
{
    int i,l,r,mid,ans;
    double a,b,ha;
    cin>>a>>b;
    a=a/100.0;
    b=b/100.0;
    for(i=0;;i++)
    {
        l=1;
        r=i;
        ans=0;
        while(l<=r)
        {
            mid=(l+r)/2;
            ha=double(mid)/double(i);
            if(ha>a)
             {
                r=mid-1;
                ans=mid;
             }
            else
             l=mid+1;
        }
        ha=double(ans)/double(i);
        if(ha<b)
         {
            cout<<i;
            return 0;
         }
    }
}

Pluto. @ 2024-10-03 15:18:50

二分法的解法是错的。我用p=13,q=14.1打了个表，发现这道题的答案没有单调性。所以你只能从最小的人数开始暴力枚举，一直枚举到ceil(100/0.01)=1e4，如果发现某一个人数可以让答案有效，直接跳出循环就好

附打表的前20个有效项：[15, 22, 23, 29, 30, 36, 37, 38, 43, 44, 45, 46, 50, 51, 52, 53, 57, 58, 59, 60]，很明显有效项是没有单调性的，所以不能用二分做

1

Sky1231 (sky1231) LV 10 @ 2021-03-16 09:49:47

#include <cmath>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <algorithm>
#include <vector>
#include <deque>
using namespace std;

namespace dts
{
    int ans;
    double p,q;
    
    void main()
    {
        scanf("%lf%lf",&p,&q);
        p/=100,q/=100;
        for (int i=1,flag=0;!flag;i++)
        {
            int l,r;
            for (l=0,r=i;r-l>1;)
            {
                int mid=(l+r)>>1;
                double rate=double(mid)/double(i);
                if (rate<=p)
                    l=mid;
                else
                    r=mid;
            }
            double rate=double(l)/double(i);
            if (p<rate&&rate<q)
            {
                flag=1;
                ans=i;
            }
            else
            {
                rate=double(r)/double(i);
                if (p<rate&&rate<q)
                {
                    flag=1;
                    ans=i;
                }
            }
        }
        printf("%d\n",ans);
    }
};

int main()
{
    dts::main();
}

1

ZeroZerg LV 8 @ 2020-06-17 16:01:31

因为人数最多10000，所以可以直接枚举，然后判断是否能够满足范围，使得最小概率乘以人数上取整可以比最大概率乘以人数下取整小于等于，但是这个算法可能会有精度问题。

但是，我要说的是，太坑了，同一套逻辑，JAVA在第六个点会出现精度问题，WA，但是CPP可以通过。
气冷抖，JAVA什么时候才能真正站起来？！！

JAVA

状态耗时内存占用

#1 Accepted 100ms 13.586 MiB
#2 Accepted 100ms 13.707 MiB
#3 Accepted 95ms 12.031 MiB
#4 Accepted 94ms 12.309 MiB
#5 Accepted 98ms 13.688 MiB
#6 Wrong Answer 104ms 13.668 MiB
#7 Accepted 97ms 13.602 MiB
#8 Accepted 97ms 13.797 MiB
#9 Accepted 97ms 12.066 MiB
#10 Accepted 103ms 12.395 MiB

CPP

状态耗时内存占用

#1 Accepted 1ms 256.0 KiB
#2 Accepted 1ms 256.0 KiB
#3 Accepted 1ms 256.0 KiB
#4 Accepted 1ms 256.0 KiB
#5 Accepted 1ms 256.0 KiB
#6 Accepted 1ms 256.0 KiB
#7 Accepted 1ms 256.0 KiB
#8 Accepted 1ms 256.0 KiB
#9 Accepted 1ms 256.0 KiB
#10 Accepted 1ms 256.0 KiB

1

lixintong LV 10 @ 2014-01-27 11:36:19

-0.000001就是考虑到i*p/100为整数的情况啦，这时的那个整数是不能取的。

1

ouou4516 LV 7 @ 2012-08-05 22:18:53

枚举整数i，直到int(q*i/100-0.00000001)-int(p*i/100+0.00000001)>=1的时候退出即可

0

GO$H LV 5 @ 2017-10-15 16:16:12

Accepted
/usr/bin/ld.bfd: warning: /out/link.res contains output sections; did you forget -T?

状态耗时内存占用

#1 Accepted 1ms 256.0 KiB
#2 Accepted 1ms 256.0 KiB
#3 Accepted 1ms 256.0 KiB
#4 Accepted 2ms 256.0 KiB
#5 Accepted 1ms 256.0 KiB
#6 Accepted 2ms 256.0 KiB
#7 Accepted 1ms 256.0 KiB
#8 Accepted 1ms 256.0 KiB
#9 Accepted 1ms 256.0 KiB
#10 Accepted 1ms 256.0 KiB
代码
var p,q,f1,f2:double;
s:longint;
begin
readln(p,q);
s:=1;
while true do
begin
f1:=s*p/100;
f2:=s*q/100;
if(f1<trunc(f2))and(f2>trunc(f2))then
begin
writeln(s);
halt;
end;
inc(s);
end;
end.

0

guhaowangjun LV 7 @ 2016-02-23 13:13:44

题目 P1041 神风堂人数
递交时间 2014-08-01 18:19:24
代码语言 C++
评测机 VijosEx
消耗时间 30 ms
消耗内存 272 KiB
评测时间 2014-08-01 18:19:28
评测结果
编译成功

测试数据 #0: Accepted, time = 0 ms, mem = 272 KiB, score = 10
测试数据 #1: Accepted, time = 0 ms, mem = 268 KiB, score = 10
测试数据 #2: Accepted, time = 0 ms, mem = 272 KiB, score = 10
测试数据 #3: Accepted, time = 0 ms, mem = 268 KiB, score = 10
测试数据 #4: Accepted, time = 15 ms, mem = 268 KiB, score = 10
测试数据 #5: Accepted, time = 0 ms, mem = 272 KiB, score = 10
测试数据 #6: Accepted, time = 0 ms, mem = 272 KiB, score = 10
测试数据 #7: Accepted, time = 0 ms, mem = 272 KiB, score = 10
测试数据 #8: Accepted, time = 15 ms, mem = 268 KiB, score = 10
测试数据 #9: Accepted, time = 0 ms, mem = 272 KiB, score = 10
Accepted, time = 30 ms, mem = 272 KiB, score = 100

0

guhaowangjun LV 7 @ 2016-02-23 13:13:42

题目 P1041 神风堂人数
递交时间 2014-08-01 18:19:24
代码语言 C++
评测机 VijosEx
消耗时间 30 ms
消耗内存 272 KiB
评测时间 2014-08-01 18:19:28
评测结果
编译成功

测试数据 #0: Accepted, time = 0 ms, mem = 272 KiB, score = 10
测试数据 #1: Accepted, time = 0 ms, mem = 268 KiB, score = 10
测试数据 #2: Accepted, time = 0 ms, mem = 272 KiB, score = 10
测试数据 #3: Accepted, time = 0 ms, mem = 268 KiB, score = 10
测试数据 #4: Accepted, time = 15 ms, mem = 268 KiB, score = 10
测试数据 #5: Accepted, time = 0 ms, mem = 272 KiB, score = 10
测试数据 #6: Accepted, time = 0 ms, mem = 272 KiB, score = 10
测试数据 #7: Accepted, time = 0 ms, mem = 272 KiB, score = 10
测试数据 #8: Accepted, time = 15 ms, mem = 268 KiB, score = 10
测试数据 #9: Accepted, time = 0 ms, mem = 272 KiB, score = 10
Accepted, time = 30 ms, mem = 272 KiB, score = 100

0

guhaowangjun LV 7 @ 2016-02-23 13:13:40

题目 P1041 神风堂人数
递交时间 2014-08-01 18:19:24
代码语言 C++
评测机 VijosEx
消耗时间 30 ms
消耗内存 272 KiB
评测时间 2014-08-01 18:19:28
评测结果
编译成功

测试数据 #0: Accepted, time = 0 ms, mem = 272 KiB, score = 10
测试数据 #1: Accepted, time = 0 ms, mem = 268 KiB, score = 10
测试数据 #2: Accepted, time = 0 ms, mem = 272 KiB, score = 10
测试数据 #3: Accepted, time = 0 ms, mem = 268 KiB, score = 10
测试数据 #4: Accepted, time = 15 ms, mem = 268 KiB, score = 10
测试数据 #5: Accepted, time = 0 ms, mem = 272 KiB, score = 10
测试数据 #6: Accepted, time = 0 ms, mem = 272 KiB, score = 10
测试数据 #7: Accepted, time = 0 ms, mem = 272 KiB, score = 10
测试数据 #8: Accepted, time = 15 ms, mem = 268 KiB, score = 10
测试数据 #9: Accepted, time = 0 ms, mem = 272 KiB, score = 10
Accepted, time = 30 ms, mem = 272 KiB, score = 100

0

guhaowangjun LV 7 @ 2016-02-23 13:13:39

题目 P1041 神风堂人数
递交时间 2014-08-01 18:19:24
代码语言 C++
评测机 VijosEx
消耗时间 30 ms
消耗内存 272 KiB
评测时间 2014-08-01 18:19:28
评测结果
编译成功

测试数据 #0: Accepted, time = 0 ms, mem = 272 KiB, score = 10
测试数据 #1: Accepted, time = 0 ms, mem = 268 KiB, score = 10
测试数据 #2: Accepted, time = 0 ms, mem = 272 KiB, score = 10
测试数据 #3: Accepted, time = 0 ms, mem = 268 KiB, score = 10
测试数据 #4: Accepted, time = 15 ms, mem = 268 KiB, score = 10
测试数据 #5: Accepted, time = 0 ms, mem = 272 KiB, score = 10
测试数据 #6: Accepted, time = 0 ms, mem = 272 KiB, score = 10
测试数据 #7: Accepted, time = 0 ms, mem = 272 KiB, score = 10
测试数据 #8: Accepted, time = 15 ms, mem = 268 KiB, score = 10
测试数据 #9: Accepted, time = 0 ms, mem = 272 KiB, score = 10
Accepted, time = 30 ms, mem = 272 KiB, score = 100

0

nzhtl1477 LV 10 @ 2014-09-08 21:05:48

P1041神风堂人数
Accepted
记录信息
评测状态 Accepted
题目 P1041 神风堂人数
递交时间 2014-09-08 21:04:44
代码语言 C++
评测机上海红茶馆
消耗时间 45 ms
消耗内存 564 KiB
评测时间 2014-09-08 21:04:45
评测结果

编译成功

测试数据 #0: Accepted, time = 0 ms, mem = 564 KiB, score = 10

测试数据 #1: Accepted, time = 0 ms, mem = 556 KiB, score = 10

测试数据 #2: Accepted, time = 0 ms, mem = 556 KiB, score = 10

测试数据 #3: Accepted, time = 0 ms, mem = 556 KiB, score = 10

测试数据 #4: Accepted, time = 0 ms, mem = 556 KiB, score = 10

测试数据 #5: Accepted, time = 15 ms, mem = 556 KiB, score = 10

测试数据 #6: Accepted, time = 0 ms, mem = 560 KiB, score = 10

测试数据 #7: Accepted, time = 0 ms, mem = 560 KiB, score = 10

测试数据 #8: Accepted, time = 15 ms, mem = 564 KiB, score = 10

测试数据 #9: Accepted, time = 15 ms, mem = 556 KiB, score = 10

Accepted, time = 45 ms, mem = 564 KiB, score = 100
代码

#include <iostream>
#include <stdio.h>
#include <string.h>
#include <algorithm>
#include <math.h>

using namespace std;

double p , q;
double a , b;
int c;
int i;

int main()
{
while( cin >> p >> q )
{
for( i = 1 ; ; i++ )
{
a = p * i / 100;
b = q * i / 100;
if( ( q * i /100 ) == floor( q * i /100 ) )
c = q * i /100 - 1;
else
c = floor( q * i / 100 );
//cout << a << " " << b << " " << c << endl;
if( a < c && b > c )
break;
}
cout << i << endl;
}
return 0;
}

用double代替float，不用考虑精度都可以。。。

0

wang_yanheng LV 10 @ 2014-08-05 14:37:32

#include <stdio.h>
#include <math.h>
#define accuracy -0.00000001
int main(){
double bottom,top;
long i;
scanf("%lf %lf",&bottom,&top);
bottom/=100.0;
top/=100.0;
for(i=1;floor(bottom*i)+1-top*i>=accuracy;i++)
;//nothing to do in this loop.
printf("%ld\n",i);
return 0;
}

0

hy__345 LV 8 @ 2014-08-01 18:20:20

1次

记录信息
评测状态 Accepted
题目 P1041 神风堂人数
递交时间 2014-08-01 18:19:24
代码语言 C++
评测机 VijosEx
消耗时间 30 ms
消耗内存 272 KiB
评测时间 2014-08-01 18:19:28
评测结果
编译成功

测试数据 #0: Accepted, time = 0 ms, mem = 272 KiB, score = 10
测试数据 #1: Accepted, time = 0 ms, mem = 268 KiB, score = 10
测试数据 #2: Accepted, time = 0 ms, mem = 272 KiB, score = 10
测试数据 #3: Accepted, time = 0 ms, mem = 268 KiB, score = 10
测试数据 #4: Accepted, time = 15 ms, mem = 268 KiB, score = 10
测试数据 #5: Accepted, time = 0 ms, mem = 272 KiB, score = 10
测试数据 #6: Accepted, time = 0 ms, mem = 272 KiB, score = 10
测试数据 #7: Accepted, time = 0 ms, mem = 272 KiB, score = 10
测试数据 #8: Accepted, time = 15 ms, mem = 268 KiB, score = 10
测试数据 #9: Accepted, time = 0 ms, mem = 272 KiB, score = 10
Accepted, time = 30 ms, mem = 272 KiB, score = 100

0

terry182 LV 8 @ 2014-06-20 23:31:10

#include <cstdio>
int main()
{ double p, q;
scanf("%lf%lf", &p, &q);
p/=100; q/=100;
for (int i = 0;;i++)
if (int(q*i-0.000001) - int(p*i+0.000001) >= 1)
{ printf("%d\n", i);
return 0;
}
return 0;
}
出現了，0ms

0

2000F11Y10M LV 8 @ 2014-05-08 21:50:53

#include <iostream>
using namespace std;
int main(){
int i;
float p,q;
cin>>p>>q;
p/=100;
q/=100;
for (i=1;i<=32767;i++){
if (int (q*i-0.000001)-int (p*i+0.000001)>=1){
cout<<i;
break;
}
}
return 0;
}

0

lingmingxiao2002 LV 8 @ 2014-04-17 19:02:05

第140道纪念

0

niyinchen LV 9 @ 2014-03-21 19:03:36

水 70道纪念特此发表

var p,q,f1,f2:double;

s:longint;

begin

readln(p,q);

s:=1;

while true do

begin
f1:=s*p/100;

f2:=s*q/100;
if(f1<trunc(f2))and(f2>trunc(f2))then

begin

writeln(s);

halt;

end;

inc(s);

end;

end.

0

0912230011 LV 8 @ 2013-11-02 20:40:43

为什么要+-0.000001

0

S.Y.F LV 10 @ 2013-09-23 20:04:16

好奇怪啊，real改成extended就能过。哪位仁兄能讲下原因？

题解

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