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总是有人喜欢show off自己的程序,说方法不就可以了么In Reply To:0MS 的程序,运用辗转相除法联立方程直接求解,最多一层循环 Posted by:2_080310236 at 2006-03-04 21:58:46 > #include "iostream.h"
> #include "math.h"
>
> void Get_Special_Solution(int a,int b,int N,int x[2])
> {
> int x1[2]={0},x2[2]={0},temp,flag1=1,flag2=1;
> if(N<0) {a=-a;b=-b;N=-N;}
> if(N==0) {x[0]=-b,x[1]=a;return;}
> if(abs(a)<abs(b)) {temp=a;a=b;b=temp;}
> if(a>0 && b<0) {b=-b;flag2=-1;}
> else if(a<0 && b>0) {a=-a;flag1=-1;}
> else if(a<0 && b<0) {a=-a;b=-b;}
> int m=a,n=b,q1,q2,r1,r2,temp1,temp2;
> q1=a/b,r1=a%b;q2=n/r1;r2=b%r1;
> m=r1;n=r2;
> x1[0]=1, x1[1]=-q1; //r1=a-q1*b;
> x2[0]=-q2, x2[1]=q1*q2+1; //r2=b-q2*r1;两个递推式来实现辗转相除法的求解过程
> if(!r2) {x[0]=flag1*x1[0];x[1]=flag2*x1[1];return;}
> while(r2!=1)
> {
> temp1=x1[0];temp2=x1[1];
> x1[0]=x2[0];x1[1]=x2[1];
> q2=m/n;r2=m%n;
> x2[0]=-q2*x2[0]+temp1;
> x2[1]=-q2*x2[1]+temp2;
> m=n;n=r2;
> }
> x[0]=flag1*x2[0];x[1]=flag2*x2[1];
> x[0]*=N;x[1]*=N;
> }
>
> void Show_Equation_Solve(int a[])
> {
> int x[2],y[2],z[2],k1,k2;
> Get_Special_Solution(28,-23,a[0]-a[1],x);
> while(x[0]>0 && x[1]>0) {x[0]-=23;x[1]-=28;}
> while(x[0]<=0||x[1]<=0) {x[0]+=23;x[1]+=28;}
> Get_Special_Solution(33,-28,a[1]-a[2],y);
> while(y[0]>0 && y[1]>0) {y[0]-=28;y[1]-=33;}
> while(y[0]<=0||y[1]<=0) {y[0]+=28;y[1]+=33;}
> if(x[0]!=y[1]) Get_Special_Solution(33,-23,x[0]-y[1],z);
> while(z[0]>0 && z[1]>0) {z[0]-=23;z[1]-=33;}
> while(z[0]<=0||z[1]<=0) {z[0]+=23;z[1]+=33;}
> a[0]=28*z[1]+x[1],a[1]=23*z[1]+x[0],a[2]=28*z[0]+y[0];
> }
>
> void main()
> {
> int i=0,m,days;
> int a[4];
> while(1)
> {
> cin>>a[0]>>a[1]>>a[2]>>a[3];
> if(a[0]==a[1] && a[1]==a[2] && a[2]==a[3] && a[3]==-1) break;
> m=a[0];
> Show_Equation_Solve(a);
> cout<<"Case "<<++i<<": the next triple peak occurs in "<<(a[0]*23+21252+m-a[3]-1)%21252+1<<" days."<<endl;
> }
> }
>
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