Parity check
Time Limit: 2000 ms
Memory Limit: 524288 KiB
Problem Description
Fascinated with the computer games, Gabriel even forgets to study. Now she needs to finish her homework, and there is an easy problem:
f(n)=
She is required to calculate f(n) mod 2 for each given n. Can you help her?
Input
Multiple test cases. Each test case is an integer n(0≤n≤
) in a single line.
Output
For each test case, output the answer of f(n)mod2.
Sample Input
2
Sample Output
1
Hint
Source
“浪潮杯”山东省第八届ACM大学生程序设计竞赛（感谢青岛科技大学）
题解:
#include<bits/stdc++.h>
using namespace std;
int main()
{
char str[1005];
while(scanf("%s",str)!=EOF)
{
int len=strlen(str);
int sum = 0,pre = 0;
for(int i = 0;i<len;i++)
{
sum = sum+(str[i]-'0');
}
if(sum%3==0)printf("0\n");
else printf("1\n");
}
return 0;
}
sum of power
Time Limit: 1000 ms
Memory Limit: 65536 KiB
Problem Description
Calculate 
mod (1000000000+7) for given n，m.
Input
Input contains two integers n,m(1≤n≤1000,0≤m≤10).
Output
Output the answer in a single line.
Sample Input
10 0
Sample Output
10
Hint
Source
“浪潮杯”山东省第八届ACM大学生程序设计竞赛（感谢青岛科技大学）
代码:
#include<bits/stdc++.h>
using namespace std;
const long long mod = 1000000000+7;
long long quik_pow(long long a,long long m)
{
long long ans = 1;
while(m)
{
if(m%2==1) ans = (ans * a) % mod ;
m = m/2 ;
a = (a*a)%mod ;
}
return ans ;
}
int main()
{
long long n,m;
scanf("%lld%lld",&n,&m);
long long sum = 0;
for(long long i =1;i<=n;i++){
sum+=quik_pow(i,m);
sum = sum%mod;
}
cout<<sum<<endl;
return 0;
}