hdu5832 A water problem (大数求余)

hdu5832 A water problem (大数求余):http://acm.hdu.edu.cn/showproblem.php?pid=5832

题目描述：

合训练的同学们~

A water problem Time Limit: 5000/2500 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others) Total Submission(s): 365    Accepted Submission(s): 201 Problem Description Two planets named Haha and Xixi in the universe and they were created with the universe beginning. There is 73 days in Xixi a year and 137 days in Haha a year.  Now you know the days N after Big Bang, you need to answer whether it is the first day in a year about the two planets.   Input There are several test cases(about 5 huge test cases). For each test, we have a line with an only integer N(0≤N), the length of N is up to 10000000.   Output For the i-th test case, output Case #i: , then output "YES" or "NO" for the answer.   Sample Input 10001 0 333   Sample Output Case #1: YES Case #2: YES Case #3: NO

题目大意：

输入一个数n，n为一个大数，若n既是73的倍数又是137的倍数，则输出YES，否则输出NO。

题目分析：

由于73和137都是质数，所以，若n为73*137的倍数，则输出YES，否则输出NO。但是由于n为大数，10的7次方以内的数，则直接开10的7次方的char数组进行输入，还要把char数组中的内容对应到int数组中，会内存超限，同样用java语言交的话，也会内存超限，为了不超限，有两种处理的方法。

代码实现一：

#include <cstdio>
#include <cstring>

const int N=1000010;
typedef long long LL ;
LL a
={0};
char a1[N*10];

int main()
{
int i,j,k;
LL d,b;
int T=0;
while(scanf("%s",a1)!=-1)
{
T++;
//if (strcmp(a1,"0")==0) {printf("Case #%d: YES\n",T);continue;}
d=0;
k=strlen(a1);
//printf("kkk%d\n",k);
for(i=0,j=0; i<k; j++)
{
if (i+10<=k)
{
a[j]=(LL)(a1[k-i-1]-'0')+(LL)(a1[k-i-2]-'0')*10+((LL)a1[k-i-3]-'0')*100+(LL)(a1[k-i-4]-'0')*1000+(LL)(a1[k-i-5]-'0')*10000+(LL)(a1[k-i-6]-'0')*100000+(LL)(a1[k-i-7]-'0')*1000000+(LL)(a1[k-i-8]-'0')*10000000+(LL)(a1[k-i-9]-'0')*100000000+(LL)(a1[k-i-10]-'0')*1000000000;
i+=10;
}
else if (i+10>k && i+9<=k)
{
a[j]=(LL)(a1[k-i-1]-'0')+(LL)(a1[k-i-2]-'0')*10+(LL)(a1[k-i-3]-'0')*100+(LL)(a1[k-i-4]-'0')*1000+(LL)(a1[k-i-5]-'0')*10000+(LL)(a1[k-i-6]-'0')*100000+(LL)(a1[k-i-7]-'0')*1000000+(LL)(a1[k-i-8]-'0')*10000000+(LL)(a1[k-i-9]-'0')*100000000;
i+=9;
}
else if (i+9>k && i+8<=k)
{
a[j]=(LL)(a1[k-i-1]-'0')+(LL)(a1[k-i-2]-'0')*10+(LL)(a1[k-i-3]-'0')*100+(LL)(a1[k-i-4]-'0')*1000+(LL)(a1[k-i-5]-'0')*10000+(LL)(a1[k-i-6]-'0')*100000+(LL)(a1[k-i-7]-'0')*1000000+(LL)(a1[k-i-8]-'0')*10000000;
i+=8;
}
else if (i+8>k && i+7<=k)
{
a[j]=(LL)(a1[k-i-1]-'0')+(LL)(a1[k-i-2]-'0')*10+(LL)(a1[k-i-3]-'0')*100+(LL)(a1[k-i-4]-'0')*1000+(LL)(a1[k-i-5]-'0')*10000+(LL)(a1[k-i-6]-'0')*100000+(LL)(a1[k-i-7]-'0')*1000000;
i+=7;
}
else if (i+7>k && i+6<=k)
{
a[j]=(LL)(a1[k-i-1]-'0')+(LL)(a1[k-i-2]-'0')*10+(LL)(a1[k-i-3]-'0')*100+(LL)(a1[k-i-4]-'0')*1000+(LL)(a1[k-i-5]-'0')*10000+(LL)(a1[k-i-6]-'0')*100000;
i+=6;
}
else if (i+6>k && i+5<=k)
{
a[j]=(LL)(a1[k-i-1]-'0')+(LL)(a1[k-i-2]-'0')*10+(LL)(a1[k-i-3]-'0')*100+(LL)(a1[k-i-4]-'0')*1000+(LL)(a1[k-i-5]-'0')*10000;
i+=5;
}
else if (i+5>k && i+4<=k)
{
a[j]=(LL)(a1[k-i-1]-'0')+(LL)(a1[k-i-2]-'0')*10+(LL)(a1[k-i-3]-'0')*100+(LL)(a1[k-i-4]-'0')*1000;
i+=4;
}
else if (i+4>k && i+3<=k)
{
a[j]=(LL)(a1[k-i-1]-'0')+(LL)(a1[k-i-2]-'0')*10+(LL)(a1[k-i-3]-'0')*100;
i+=3;
}
else if (i+3>k && i+2<=k)
{
a[j]=(LL)(a1[k-i-1]-'0')+(LL)(a1[k-i-2]-'0')*10;
i+=2;
}
else if (i+2>k && i+1<=k)
{
a[j]=(LL)(a1[k-i-1]-'0');
i++;
}
}
// for(int i=j-1; i>=0; i--)
// {
// printf("%I64d ",a[i]);
// }
// printf("\n");
for(i=j-1; i>=0; i--)
{
d=d*10000000000+a[i];
d=d%10001;
}
if (d==0) printf("Case #%d: YES\n",T);
else printf("Case #%d: NO\n",T);
}
return 0;
}

代码实现二：
#include <iostream>
#include <cstdio>
#include <cstring>

using namespace std;

const int mod=10001;

char str[10000100];

int main()
{
long long sum;
int casenum=1;
while(scanf("%s",str)!=EOF)
{
sum=0;
int len=strlen(str);
for(int i=0;i<len;i++)
{
sum=sum*10+(str[i]-'0');
sum=sum%mod;
}
if(sum==0)
{
printf("Case #%d: YES\n",casenum++);
}
else
{
printf("Case #%d: NO\n",casenum++);
}
}
return 0;
}