HDU-5832 HDU-1212/2016网络选拔赛/大数取余
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A water problem
Time Limit: 5000/2500 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)Total Submission(s): 326 Accepted Submission(s): 180
Problem Description
Two planets named Haha and Xixi in the universe and they were created with the universe beginning.
There is73 days in Xixi a year and 137 days in Haha a year.
Now you know the daysN after Big Bang, you need to answer whether it is the first day in a year about the two planets.
There is
Now you know the days
Input
There are several test cases(about 5 huge test cases).
For each test, we have a line with an only integerN(0≤N) , the length of N is up to 10000000 .
For each test, we have a line with an only integer
Output
For the i-th test case, output Case #i: , then output "YES" or "NO" for the answer.
Sample Input
100010333
Sample Output
Case #1: YESCase #2: YESCase #3: NO
Author
UESTC
Source
2016中国大学生程序设计竞赛 - 网络选拔赛
Cmod(A*B)==0 =CmodA==0&&CmodB==0
C
---------- =integer
A*B
#include <iostream>
using namespace std;
#define N 12000000
#define INF 1<<30
char s[N];
int main()
{
int k=0,len,x,d,i;
while(scanf("%s", s) != EOF)
{
k++;
len = strlen(s);
x=10001;
d=0;
for(i=0;i<len;i++)
d=(d*10+(s[i]-'0')%x)%x;
using namespace std;
#define N 12000000
#define INF 1<<30
char s[N];
int main()
{
int k=0,len,x,d,i;
while(scanf("%s", s) != EOF)
{
k++;
len = strlen(s);
x=10001;
d=0;
for(i=0;i<len;i++)
d=(d*10+(s[i]-'0')%x)%x;
if(d==0)
printf("Case #%d: YES\n",k);
else
printf("Case #%d: NO\n",k);
}
return 0;
}
printf("Case #%d: YES\n",k);
else
printf("Case #%d: NO\n",k);
}
return 0;
}
Big Number
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 7409 Accepted Submission(s): 5124
Problem Description
As we know, Big Number is always troublesome. But it's really important in our ACM. And today, your task is to write a program to calculate A mod B.
To make the problem easier, I promise that B will be smaller than 100000.
Is it too hard? No, I work it out in 10 minutes, and my program contains less than 25 lines.
To make the problem easier, I promise that B will be smaller than 100000.
Is it too hard? No, I work it out in 10 minutes, and my program contains less than 25 lines.
Input
The input contains several test cases. Each test case consists of two positive integers A and B. The length of A will not exceed 1000, and B will be smaller than 100000. Process to the end of file.
Output
For each test case, you have to ouput the result of A mod B.
Sample Input
2 312 7152455856554521 3250
Sample Output
251521
Author
Ignatius.L
Source
杭电ACM省赛集训队选拔赛之热身赛
#include <iostream>
#include<string.h>
using namespace std;
#define N 1005
char s[N];
int main()
{
int k=0,len,x,d,i;
while(scanf("%s",s)!= EOF) {
cin>>x;
k++;
len = strlen(s);
d=0;
cin>>x;
k++;
len = strlen(s);
d=0;
for(i=0;i<len;i++)
d=(d*10+(s[i]-'0')%x)%x;
printf("%d\n",d);
d=(d*10+(s[i]-'0')%x)%x;
printf("%d\n",d);
}
return 0;
}
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