hdu 1788

题目链接 http://acm.hdu.edu.cn/showproblem.php?pid=1788

其实有两种解法一种是转化为求同余方程组然后再解，第二种方法是根据同余定理推出 N+a≡0（mod Mi）有了这个式子，就知道这个题目意思就是求N个Mi的最小公倍数了

 1 #include <cstdio>
 2 #include <cstring>
 3 #include <iostream>
 4 #include <stack>
 5 #include <queue>
 6 #include <map>
 7 #include <algorithm>
 8 #include <vector>
 9 
10 using namespace std;
11 
12 const int maxn = 1000005;
13 
14 typedef long long LL;
15 
16 LL ex_gcd(LL a,LL b,LL &x,LL &y)
17 {
18    if(b == 0){
19         x = 1;
20         y = 0;
21         return a;
22    }
23    LL r = ex_gcd(b,a%b,x,y);
24    LL t = x;
25       x = y;
26       y = t - a/b*y;
27       return r;
28 }
29 int main()
30 {
31     LL i,n,a1,aa,r1,a2,r2,ans,a,b,c,d,x0,y0;
32     while(scanf("%lld%lld",&n,&aa)!=EOF){
33         bool flag = 1;
34         if(n == 0 && aa==0) break;
35         scanf("%lld",&a1);
36         r1 = a1-aa;
37         for( i=1;i<n;i++){
38             scanf("%lld",&a2);
39             r2 = a2 - aa;
40             a = a1;
41             b = a2;
42             c = r2-r1;
43             LL d = ex_gcd(a,b,x0,y0);
44 
45             int t = b/d;
46             x0 = (x0*(c/d)%t+t)%t;
47             r1 = a1*x0 + r1;
48             a1 = a1*(a2/d);
49 
50         }
51 
52         printf("%lld\n",r1);
53     }
54     return 0;
55 }
View Code
 1 #include <cstdio>
 2 #include <cstring>
 3 #include <iostream>
 4 #include <stack>
 5 #include <queue>
 6 #include <map>
 7 #include <algorithm>
 8 #include <vector>
 9 
10 using namespace std;
11 
12 const int maxn = 1000005;
13 
14 typedef long long LL;
15 
16 LL gcd( LL a,LL b)
17 {
18     if(b == 0) return a;
19    else  return gcd(b,a%b);
20 }
21 int main()
22 {
23     LL i,n,a1,aa,r1,a2,r2,ans,a,b,c,d,x0,y0,lcm;
24     while(scanf("%lld%lld",&n,&aa)!=EOF){
25         bool flag = 1;
26         if(n == 0 && aa==0) break;
27         lcm = 1;
28         for(int i=0;i<n;i++){
29             cin>>a;
30             lcm = (lcm*a)/gcd(lcm,a);
31         }
32         printf("%lld\n",lcm-aa);
33     }
34     return 0;
35 }
View Code

 

posted @ 2015-11-03 22:07  lmlyzxiao  阅读(179)  评论(0)  编辑  收藏  举报