poj 2891 Strange Way to Express Integers 中國剩餘定理

Strange Way to Express Integers

Time Limit: 1000MS		Memory Limit: 131072K
Total Submissions: 4191		Accepted: 1100

Description

 

Elina is reading a book written by Rujia Liu, which introduces a strange
way to express non-negative integers. The way is described as following:

 

Choose k different positive integers a1,
a2, …, ak. For some
non-negative m, divide it by every ai (1 ≤ i ≤
k) to find the remainder ri. If a1,
a2, …, ak are properly chosen, m can be
determined, then the pairs (ai, ri) can be
used to express m.

“It is easy to calculate the pairs from m, ” said Elina. “But how can
I find m from the pairs?”

Since Elina is new to programming, this problem is too difficult for her. Can
you help her?

Input

The input contains multiple test cases. Each test cases consists of some
lines.

• Line 1: Contains the integer k.
• Lines 2 ~ k + 1: Each contains a pair of integers
  ai, ri (1 ≤ i ≤
  k).

 

Output

Output the non-negative integer m on a separate line for each test
case. If there are multiple possible values, output the smallest one. If there
are no possible values, output -1.

 

Sample Input

28 711 9

Sample Output

31

Hint

All integers in the input and the output are non-negative and can be
represented by 64-bit integral types.

Source

POJ
Monthly--2006.07.30, Static

 

/*<br />對於w[1],w[2].....w[n]不互素的情形，就只能兩個兩個來求了<br />x=a[1] (mod m[1])<br />x=a[2] (mod m[2])<br />解完後，a=x,m=m1和m2的最小公倍數<br />將題目意思轉化為公式：a1*x-a2*y=r2-r1,用歐幾裡得擴充演算法求解</p><p>a1+m1*y=a2+m2*z ==> a1-a2=m2*z-m1*y<br />則(a1-a2)%gcd(m1,m2)=0<br />則:x=a2+m2*z (1)<br />（應該a>m，必需條件）<br />x=r1 (mod a1)<br />x=r2 (mod a2)<br />---> //a1*x+a2*y=gcd(a1,a2)=d<br />---> //r1+a1*x=r2+a2*y ---> a1*x+a2*y=r2-r1<br />則(r2-r1)%d!=0 不能得出<br />否則<br />{<br /> t=a2/d;<br /> x=((x*(r2-r1)/d)%t+t)%t//x的最小整數解<br /> r1=x*a1+r1 //由（1）式<br /> a1=a1*a2/d;<br /> r1=(r1%a1+a1)%a1;<br />}<br />*/<br />#include<cstdio><br />#include<cstdlib><br />long long int ex_Gcd(long long int p,long long int q,long long int &x,long long &y)<br />{<br /> long long int temp,r;<br /> if(q==0)<br /> {<br /> x=1;<br /> y=0;<br /> return p;<br /> }<br /> r=ex_Gcd(q,p%q,x,y);<br /> temp=x;<br /> x=y;<br /> y=temp-p/q*y;<br /> return r;<br />}<br />int main()<br />{<br /> long long int n,a1,a2,r1,r2,x,y,i,d,t;<br /> bool flag;<br /> while(scanf("%lld",&n)!=EOF)<br /> {<br /> scanf("%lld %lld",&a1,&r1);<br /> if(n==1)<br /> {<br /> if(a1<=r1) //餘數不可能大於除數<br /> printf("-1/n");<br /> else<br /> printf("%lld/n",r1);<br /> continue;<br /> }<br /> flag=true;<br /> for(i=1; i<n; i++)<br /> {<br /> scanf("%lld %lld",&a2,&r2);<br /> if(!flag) continue;<br /> if(a2<=r2)<br /> flag=false;<br /> d=ex_Gcd(a1,a2,x,y);//a1*x+a2*y=gcd(a1,a2)<br /> if((r2-r1)%d!=0) flag=false;<br /> else<br /> {<br /> t=a2/d;<br /> x=((r2-r1)/d*x%t+t)%t;<br /> r1=x*a1+r1;<br /> a1=a1*a2/d; //a1為a1和a2的最小公倍數<br /> r1=(r1%a1+a1)%a1;<br /> }<br /> }</p><p> if(flag)<br /> printf("%lld/n",r1);<br /> else<br /> printf("-1/n");<br /> }<br /> return 0;<br />}<br />