Just RandomTime Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 1723 Accepted Submission(s): 483
Problem Description
Coach Pang and Uncle Yang both love numbers. Every morning they play a game with number together. In each game the following will be done:
1. Coach Pang randomly choose a integer x in [a, b] with equal probability. 2. Uncle Yang randomly choose a integer y in [c, d] with equal probability. 3. If (x + y) mod p = m, they will go out and have a nice day together. 4. Otherwise, they will do homework that day. For given a, b, c, d, p and m, Coach Pang wants to know the probability that they will go out.
Input
The first line of the input contains an integer T denoting the number of test cases.
For each test case, there is one line containing six integers a, b, c, d, p and m(0 <= a <= b <= 109, 0 <=c <= d <= 109, 0 <= m < p <= 109).
Output
For each test case output a single line "Case #x: y". x is the case number and y is a fraction with numerator and denominator separated by a slash (‘/‘) as the probability that they will go out. The fraction should be presented in the simplest form (with
the smallest denominator), but always with a denominator (even if it is the unit).
Sample Input
Sample Output
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题目
输入 z1 y1 z2 y2 q m
问 z1到y1 中取一个数 再从 z2到y2中取一个数， 两数加和 对q取模 ，其值等于m的几率是多大
很容易知道总的选择，为(y2-z2+1)*(y1-z1+1)；
然后可以通过容斥原理来算分子
dd=count(y1,y2,q,m); 这个算的是0-y1和0-y2中的任意两数和对q取模答案是m的方法数 为了好记，dd的意义是大大
xd=count(z1-1,y2,q,m); 接下来三个同上 xd意义是 大小
dx=count(z2-1,y1,q,m);
xx=count(z1-1,z2-1,q,m);
根据容斥原理，可以得到ans=dd-dx-xd+xx
dd因为是从0开始算，所以多算了，-dx-xd可以把有比z1 z2小的那些方法数给去掉，但是去的过程中又多去了 xx，所以把xx加回来
count(__int64 num1,__int64 num2,__int64 q,__int64 m)
count可以自己算，想法就是把两个输入的数按q个数拆分，num1这个数按q来分，可以分成rep1段，剩下m1个数字不足q
__int64 ji(__int64 m1,__int64 m2,__int64 q,__int64 m)
这个函数是计算 有m1个数，比如说m1=4 那第一组数字就是0，1，2，3这四个数字，第二组有m2个数，也是从0开始算的；
这两个数都是比q小的，可以用ji（）这个函数，算出符合要求的方法数；
#include<stdio.h>
__int64 ji(__int64 m1,__int64 m2,__int64 q,__int64 m) // m1 m2 是个数 m是目标
{
__int64 ans=0;
__int64 tem=q-1-m;// 最多
if(tem-(q-m1)-(q-m2)>0)
ans+=tem-(q-m1)-(q-m2);
m1--;// 从有几个数字 变成 这个数字剩下数最大多少
m2--;
//没取模
if(m1>=m&&m2>=m)
{
ans+=(m+1);
}
else if(m1>=m)
{
ans+=m2+1;
}
else if(m2>=m)
{
ans+=m1+1;
}
else
{
if(m1+m2>=m)
{
ans+=m2-(m-m1)+1;
}
}
return ans;
}
__int64 count(__int64 num1,__int64 num2,__int64 q,__int64 m)//m < p
{
__int64 rep1,rep2,ans,m1,m2;
rep1=(num1+1)/q;
rep2=(num2+1)/q;
ans=0;
ans+=rep1*rep2*(m+1);//前段不取余
if(m!=q-1)//取余的
ans+=(q-1-m)*rep1*rep2;
//尾巴和头的 要加上
m1=(num1+1)%q; // 1 - 剩下的个数
m2=(num2+1)%q;
ans+=ji(m1,m2,q,m);
ans+=rep2*ji(m1,q,q,m);
ans+=rep1*ji(q,m2,q,m);
return ans;
}
__int64 gcd(__int64 n,__int64 m){//最大公因数
__int64 tmp;
while(m){tmp=n;n=m;m=tmp%m;}
return n;
}
int main()
{
__int64 ans,yue,fenmu;
__int64 t,z1,y1,z2,y2;
__int64 q,m,dd,dx,xd,xx;
__int64 cas=1;
scanf("%I64d",&t);
while(t--)
{
/*调试用
printf("0-a 0-b 有多少 匹配 相加%q等于m\n");
scanf("%I64d%I64d%I64d%I64d",&z1,&z2,&q,&m);
printf("%I64d\n",count(z1,z2,q,m));
__int64 i,j,ans=0;
for(i=0;i<=z1;i++)
{
for(j=0;j<=z2;j++)
{
if((i+j)%q==m)
ans++;
}
}
printf("%I64d\n",ans);
*/
scanf("%I64d%I64d",&z1,&y1);
scanf("%I64d%I64d",&z2,&y2);
scanf("%I64d%I64d",&q,&m);
dd=count(y1,y2,q,m);
xd=dx=xx=0;
if(z1!=0)
xd=count(z1-1,y2,q,m);
if(z2!=0)
dx=count(z2-1,y1,q,m);
if(z1!=0&&y1!=0)
xx=count(z1-1,z2-1,q,m);
//printf("dd%I64d dx%I64d xd%I64d xx%I64d\n",dd,dx,xd,xx);
fenmu=(y2-z2+1)*(y1-z1+1);//总的
ans=dd-dx-xd+xx;
yue=gcd(ans,fenmu);
printf("Case #%I64d: ",cas++);
if(ans==0)
{
puts("0/1");
}
else
printf("%I64d/%I64d\n",ans/yue,fenmu/yue);
}
return 0;
}
Just RandomTime Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 1723 Accepted Submission(s): 483
Problem Description
Coach Pang and Uncle Yang both love numbers. Every morning they play a game with number together. In each game the following will be done:
1. Coach Pang randomly choose a integer x in [a, b] with equal probability. 2. Uncle Yang randomly choose a integer y in [c, d] with equal probability. 3. If (x + y) mod p = m, they will go out and have a nice day together. 4. Otherwise, they will do homework that day. For given a, b, c, d, p and m, Coach Pang wants to know the probability that they will go out.
Input
The first line of the input contains an integer T denoting the number of test cases.
For each test case, there is one line containing six integers a, b, c, d, p and m(0 <= a <= b <= 109, 0 <=c <= d <= 109, 0 <= m < p <= 109).
Output
For each test case output a single line "Case #x: y". x is the case number and y is a fraction with numerator and denominator separated by a slash (‘/‘) as the probability that
they will go out. The fraction should be presented in the simplest form (with the smallest denominator), but always with a denominator (even if it is the unit).
Sample Input
Sample Output
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原文地址：http://blog.csdn.net/u013532224/article/details/40984799
