A water problem
Time Limit: 5000/2500 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 97 Accepted Submission(s): 58
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
Input
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
Author
UESTC
Source
真是被这个题目恶心到了，
其实也怪自己见识的太少，用了很长的时间的高精度，最后用高精度加法过了，最后一问同学，简直是个水题！
都说一下吧，哎~~~
因为是一个大数取模吗，既是73的倍数，又是137的倍数，所以73*137 = 10001，所以可以枚举每一个数字，边枚举边取模，这样还哪能用到高精度啊，，
就像转换成10进制一样v = （v * 10 + a ) % mod，这样如果最后v是0 就yes了，否则就no！
高精度思路：
可能是找的板子有问题，全部都是超时，爆内存，
转换一下，把10001 = 10000 + 1
这就相当于输入n ，找出一个k 满足 k*10000 + k = n，这就相当于两个错位加法运算，直接模拟加法就行了！
奇怪的是，JAVA交好像没有过的~全都是爆内存！
水题代码：
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
const int maxn = 10000000 + 1;
const int mod = 10001;
char s[maxn];
int main(){
int kase = 0;
while(scanf("%s",s) == 1){
int len = strlen(s);
int v = 0;
for (int i = 0; i < len; ++i){
v = (v*10 + s[i]-48) % mod;
}
if (v == 0)printf("Case #%d: YES\n",++kase);
else printf("Case #%d: NO\n",++kase);
}
return 0;
}
/*
55894144405555
*/
找抽代码：
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
const int maxn = 10000000 + 1;
char s[maxn];
char a[maxn];
char b[maxn];
char ans[maxn];
int main(){
int kase = 0 ;
while(scanf("%s",s+1) == 1){
int len = strlen(s+1);
if (len > 4){
a[len] = s[len]-48;
a[len-1] = s[len-1]-48;
a[len-2] = s[len-2]-48;
a[len-3] = s[len-3]-48;
int flag = 0;
for (int i = 1; i <= len-4; ++i){
int shang = a[len-i+1];
int xia = s[len-4-i+1]-48;
if (xia >= flag + shang)a[len-4-i+1] = xia - shang - flag,flag = 0;
else a[len-4-i+1] = xia+10-flag-shang,flag = 1;
}
a[len+1] = a[len+2] = a[len+3] = a[len+4] = 0;
bool ok =true;
int cnt = 0;
for (int i = 1; i <= len; ++i)a[i] = a[i+4];
// for (int i = 1; i <= len; ++i)printf("%d ",a[i]);
// puts("");
// for (int i = 1; i <= len; ++i)printf("%d ",a[i]);
// puts("");
b[1] = b[2] = b[3] = b[4] = 0;
for (int i =5; i <= len; ++i)b[i] = a[i-4];
// for (int i = 1; i <= len; ++i)printf("%d ",b[i]);
// puts("");
flag = 0;
for (int i = len ;i >= 1; --i){
int t = a[i] + b[i] + flag;
if (t > 9)flag = 1,t %= 10;
else flag = 0;
ans[i] = t;
}
// for (int i = 1; i <= len ;++i)printf("%d ",ans[i]);
for (int i = 1; i <= len; ++i){
if (s[i]-48 != ans[i]){
ok = false;
break;
}
}
if (ok)printf("Case #%d: YES\n",++kase);
else printf("Case #%d: NO\n",++kase);
}else {
int v = 0 ;
for (int i = 1; i <= len; ++i){
v = v * 10 + s[i] - 48;
}
if (v % 10001 == 0)printf("Case #%d: YES\n",++kase);
else printf("Case #%d: NO\n",++kase);
}
}
return 0;
}
/*
55894144405555
*/
