Hdu5429, math + high precision + equal-ratio sequence to judge __ Math

Meaning

Give n high precision numbers to determine whether the equal ratio sequence.

Range: 0<n<=100, each value is no more than 100 length.

Analysis:

According to the question, this problem must be solved with high precision, using the array simulation multiplication calculation process.

Determine whether the equal-ratio sequence method: a[i],a[i+1],a[i+2], for arbitrary I, have a[i]*a[i+2]=a[i+1]*a[i+1].

Therefore, a high-precision simulation of the multiplication calculation, to determine whether the adjacent three items are satisfied with the above relational formula.

Attention:

The number given in the ① topic may have a leading 0;

The ② ratio of the equal-ratio sequence is nonzero; but when the sequence is all zero, it is also the equal-ratio sequence.

such as: 0 0 0 0 is the equal-ratio sequence, because the ratio can be 1,2 ...,

And: 1 0 0 0 is not the equal ratio sequence, the male ratio is 0.

Notice the two things, the specific code is relatively simple, pay attention to the high-precision multiplication of the writing details.

Specific code:

#include <cstdio> #include <algorithm> #include <iostream> #include <vector> #include <
Cstring> #include <cmath> using namespace std;
typedef long Long LL;
int t,n;
Char s[110][110],str[110][110];
int res[300];
Char res[300]; String product (char A[],char b[]) {//cout<< "***************************" <<endl;//COUT&LT;&LT;A&LT;&L t; '
    ' <<b<<endl;
    int Len1=strlen (a);
    int Len2=strlen (b);
    memset (res,0,sizeof (res));
    memset (res,0,sizeof (Res));
            for (int i=0;i<len1;i++)//2 {for (int j=0;j<len2;j++)//1 {int c,d;
            c=a[i]-' 0 ';
            d=b[j]-' 0 ';
            Res[i+j]+=c*d;
            res[i+j+1]+= (RES[I+J]/10);
            res[i+j]%=10;
        printf ("%d%d%d%d\n", c,d,res[i+j],res[i+j+1]);
    for (int i=0;i<len1+len2-1;i++) res[i]=res[i]+ ' 0 ';
if (res[len1+len2-1]) res[len1+len2-1]=res[len1+len2-1]+ ' 0 '; Cout<<res<<endl;
return Res; BOOL Judge (string A,string B) {//cout<<a<< ' <<b<<endl;/cout<< '---------------
    -----------"<<endl;
    int len1=a.size ();
    int len2=b.size ();
    if (LEN1!=LEN2) return false;
    for (int i=0;i<len1;i++) {if (A[i]!=b[i]) return false;
return true;
    int main () {scanf ("%d", &t);
        while (t--) {scanf ("%d", &n);
        for (int i=0; i<n; i++) {scanf ("%s", Str[i]);
        int e=0;
            for (int i=0;i<n;i++) {int Len=strlen (str[i]);
            for (int j=len-1;j>=0;j--) {s[i][len-1-j]=str[i][j];
            while ((s[i][len-1]== ' 0 ') && (len>1)) {len-=1;
            } s[i][len]= ';
            if ((len==1) && (s[i][len-1]== ' 0 ')) {e++; }//Cout<<s[i]<<eNdl
            } if (e>0) {if (e<n) cout<< "No" <<endl;
            else cout<< "Yes" <<endl;
        Continue
        BOOL Flag=true;
String a,b;
        Char a[200],b[200]; for (int i=1;i<n-1;i++) {a=product (s[i-1],s[i+1]);//cout<< "a=" <<A<<
            Endl
B=product (S[i],s[i]);
            cout<< "b=" <<B<<endl;
            BOOL F=judge (A,B);
            if (!f) {flag=false;
        } if (flag) cout<< "Yes" <<endl;
    else cout<< "No" <<endl;
return 0;

 }/* 10 1 0 3 1 1 1 3 1 4 2 5 16 8 4 2/1 * *