題目如下：

As a simple and boring task, you are asked to do modulo on floating numbers.

Given a, b, please calculate .

Notice that the modulo supported by some programming languages like C++, Java or Python might be not enough. You may have to implement a special version for this task.

Input
Two space-separated floating numbers a, b (0
 
 < a, b ≤ 109). Exactly 9 digits are kept after the decimal point.

Output
Print the answer with absolute or relative error not greater than 10 - 15.

Examples
Input
3.000000000 2.000000000
Output
1.000000000
Input
0.400000000 0.200000000
Output
0

題意：輸入浮點數a與b,輸出a%b的值。（a,b已經確認小數點後最多有9個數）

AC程式碼：

#include<iostream>
#include<cstdio>
#include
 
<cstring>
#include<string>
#include<cstdlib>
#include<cmath>
#include<algorithm>
#include<vector>
#include<queue>
#include<stack>
#include<set>
#define Max(a,b) ((a)>(b)?(a):(b))
#define Min(a,b) ((a)<(b)?(a):(b))
#define Swap(a,b,t) t=a,a=b,b=t
#define
 
 Mem0(x) memset(x,0,sizeof(x))
#define Mem1(x) memset(x,-1,sizeof(x))
#define MemX(x) memset(x,0x3f,sizeof(x))
using namespace std;
typedef long long ll;
const int inf=0x3f3f3f;
const double eps=1e-12;
int main()
{
    ll a,b;
    char s1[20],s2[20];
    while (scanf("%lld.%s%lld.%s",&a,s1,&b,s2)!=EOF){
        for (int i=0;i<9;i++){
               a=a*10+(int)(s1[i]-'0');
        }
        for (int j=0;j<9;j++){
            b=b*10+(int)(s2[j]-'0');
        }
        ll mod=a%b;
        long double ans;
        ans=mod/1e9;
        if (ans<eps)
            printf("0\n");
        else
            printf("%.9Lf\n",ans);        
    }
    return 0;
}