高精度求余-HDOJ-5373-The Shortest Problem

The shortest problem

Time Limit: 3000/1500 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 1676 Accepted Submission(s): 746

Problem Description
In this problem, we should solve an interesting game. At first, we have an integer n, then we begin to make some funny change. We sum up every digit of the n, then insert it to the tail of the number n, then let the new number be the interesting number n. repeat it for t times. When n=123 and t=3 then we can get 123->1236->123612->12361215.

Input
Multiple input.
We have two integer n (0<=n<=104 ) , t(0<=t<=105) in each row.
When n==-1 and t==-1 mean the end of input.

Output
For each input , if the final number are divisible by 11, output “Yes”, else output ”No”. without quote.

Sample Input
35 2
35 1
-1 -1

Sample Output
Case #1: Yes
Case #2: No

//
// main.cpp
// 150811-1005
//
// Created by 袁子涵 on 15/8/11.
// Copyright (c) 2015年 袁子涵. All rights reserved.
//

#include <iostream>
#include <string.h>
#include <stdio.h>
#include <stdlib.h>

using namespace std;

int n;
long int t;
char num2[100],num[700000];
unsigned long int lth1,lth2,lth3;
unsigned long long int ss;

bool mod11()
{
    lth1=strlen(num);
    unsigned long long int ji=0,ou=0;
    for (int i=0; i<lth1; i+=2) {
        ji+=num[i]-48;
    }
    for (int i=1; i<lth1; i+=2) {
        ou+=num[i]-48;
    }
    if (((ji-ou)>0?(ji-ou):(ou-ji))%11==0 || ji-ou==0) {
        return true;
    }
    else
        return false;
}

void intos(long long int x,char *a)
{
    char biao[10]={'0','1','2','3','4','5','6','7','8','9'};
    long int i=0;
    while (x) {
        a[i]=biao[x%10];
        i++;
        x/=10;
    }
    char temp;
    for (int j=0; j<=(i-1)/2; j++) {
        temp=a[j];
        a[j]=a[i-1-j];
        a[i-1-j]=temp;
    }
    return;
}

int main(int argc, const char * argv[]) {
    int k=1;
    while (scanf("%d %ld",&n,&t)!=EOF && n!=-1 && t!=-1) {
        if (n==0) {
            printf("Case #%d: Yes\n",k);
            k++;
            continue;
        }
        if (t==0) {
            if (n%11==0) {
                printf("Case #%d: Yes\n",k);
                k++;
                continue;
            }
            else
            {
                printf("Case #%d: No\n",k);
                k++;
                continue;
            }
        }
        memset(num, 0, sizeof(num));
        memset(num2, 0, sizeof(num2));
        intos(n, num);
        ss=0;
        lth1=n;
        while (lth1) {
            ss+=lth1%10;
            lth1/=10;
        }
        intos(ss, num2);
        strcat(num, num2);
        t--;
        while (t--) {
            lth1=strlen(num2);
            for (long long int i=0; i<lth1; i++) {
                ss+=num2[i]-48;
            }
            intos(ss, num2);
            strcat(num, num2);
        }
        if (mod11()) {
            printf("Case #%d: Yes\n",k);
        }
        else
            printf("Case #%d: No\n",k);
        k++;
    }
    return 0;
}