Strange Way to Express Integers
Time Limit:1000MSMemory Limit:131072KB64bit IO Format:%I64d & %I64u
POJ 2891
Appoint description:
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:
Choosekdifferent positive integersa1,a2,…,ak. For some non-negativem, divide it by everyai(1 ≤i≤k) to find the remainderri. Ifa1,a2, …,akare properly chosen, m can be determined, then the pairs (ai,ri) can be used to expressm.
“It is easy to calculate the pairs fromm, ” said Elina. “But how can I findmfrom 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 integerk.Lines 2 ~k+ 1: Each contains a pair of integersai,ri(1 ≤i≤k).
Output
Output the non-negative integermon 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.
题意：给你k组数。x%M[i]=A[i];
思路：中国剩余定理，扩展欧几里德
不会的可以参考：
转载请注明出处：
题目链接：?id=2891
#include<stdio.h>#define LL __int64void exgcd(LL a,LL b,LL& d,LL& x,LL& y){if(!b){d=a;x=1;y=0;}else{exgcd(b,a%b,d,y,x);y-=x*(a/b);}}LL gcd(LL a,LL b){if(!b){return a;}gcd(b,a%b);}LL M[55000],A[55000];LL China(int r){LL dm,i,a,b,x,y,d;LL c,c1,c2;a=M[0];c1=A[0];for(i=1; i<r; i++){b=M[i];c2=A[i];exgcd(a,b,d,x,y);c=c2-c1;if(c%d) return -1;//c一定是d的倍数，如果不是，则，肯定无解dm=b/d;x=((x*(c/d))%dm+dm)%dm;//保证x为最小正数//c/dm是余数，，系数扩大余数被c1=a*x+c1;a=a*dm;}if(c1==0)//余数为0，说明M[]是等比数列。且余数都为0{c1=1;for(i=0;i<r;i++)c1=c1*M[i]/gcd(c1,M[i]);}return c1;}int main(){int n;while(scanf("%d",&n)!=EOF){for(int i=0;i<n;i++){scanf("%I64d%I64d",&M[i],&A[i]);}if(n==1){ printf("%I64d\n",A[0]);continue;}LL ans=China(n);printf("%I64d\n",ans);}return 0;}
你不勇敢，没人替你坚强。
