Time Limit: 1000MS | Memory Limit: 10000K | |
Total Submissions: 138199 | Accepted: 44268 |
Description
Input
当p = e = i = d = -1时,输入数据结束。
Output
采用以下格式:
Case 1: the next triple peak occurs in 1234 days.
注意:即使结果是1天,也使用复数形式“days”。
Sample Input
0 0 0 0 0 0 0 100 5 20 34 325 4 5 6 7 283 102 23 320 203 301 203 40 -1 -1 -1 -1
Sample Output
Case 1: the next triple peak occurs in 21252 days. Case 2: the next triple peak occurs in 21152 days. Case 3: the next triple peak occurs in 19575 days. Case 4: the next triple peak occurs in 16994 days. Case 5: the next triple peak occurs in 8910 days. Case 6: the next triple peak occurs in 10789 days.
Source
Translator
x ≡ a2(mod m2)
.
.
.
.
.
.
x ≡ ak(mod mk)
的方程组.
其中 mi 两两互质.
常用于求 mn(mod p) ,其中n需要取模而p是个质数
令 M =∏ni=1 mi
.
令 Mi =M/mi
.
令 ki ≡ Mi-1(mod mi)
则模 M 意义下的唯⼀解为 ∑ni=1aikiMi
ki可用exgcd求得
不证明了也不会。。(数学渣)
#include <ctype.h> #include <cstdio> #define lcm 21252 void read(int &x) { x=0;bool f=0; register char ch=getchar(); for(;!isdigit(ch);ch=getchar()) if(ch=='-') f=1; for(; isdigit(ch);ch=getchar()) x=(x<<3)+(x<<1)+ch-'0'; x=f?(~x)+1:x; } int A[5],m[5]; void exgcd(int a,int b,int &x,int &y) { if(b==0) { x=1; y=0; return; } exgcd(b,a%b,x,y); int tmp=x; x=y; y=tmp-a/b*y; } int CRT() { int M=1; int ans=0; for(int i=1;i<=3;i++) M*=m[i]; int x,y,Mi; for(int i=1;i<=3;i++) { Mi=M/m[i]; exgcd(Mi,m[i],x,y); ans=(ans+Mi*x*A[i])%M; } if(ans<0) ans+=M; return ans; } int main() { int a,b,c,d,cas=0; for(;;) { read(a);read(b);read(c);read(d); if(a==-1&&b==-1&&c==-1&&d==-1) break; A[1]=a; A[2]=b; A[3]=c; m[1]=23; m[2]=28; m[3]=33; int ans=CRT(); if(ans<=d) ans+=lcm; printf("Case %d: the next triple peak occurs in %d days.\n",++cas,ans-d); } return 0; }