The remainder of nine

Remainder time limit of nine: theMs | Memory Limit:65535KB Difficulty:3

Describe

Now give you a natural number n, whose number is less than or equal to 1 million, now all you have to do is find the remainder after the number is divisible by nine.

Input

the first line has an integer m (1<=m<=8), which indicates that there is an M-group of test data;
The M row then has a natural number n per row.

Output

output n is divided by the remainder after nine, with each output occupying one row.

Sample input

345465456541

Sample output

454

Source

[Miao Building] Original

Uploaded by

Miao-dong Building

Review the division of large numbers ~ ~

#include <iostream> #include <stdio.h> #include <string.h>using namespace Std;char a[1000000];int Main () {int T,yushu,len,i,s;cin>>t;while (t--) {yushu=0;        cin>>a;        Len=strlen (a);       for (i=0;i<len;i++)   {   s=yushu*10+ (a[i]-' 0 ');   yushu=s%9;   }   Cout<<yushu<<endl;} return 0;}        

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The remainder of nine