The series, 11 + 22 + 33 + ... + 1010 = 10405071317.
Find the last ten digits of the series, 11 + 22 + 33 + ... + 10001000.
Go to the thread for problem 48 in the forum.
ans=0
i=1
F=10**10
for i in range(1,1001):
p=1
for j in range(i): p=p*i%F
ans=(ans+p)%F
print i
print ans
The prime factors of 13195 are 5, 7, 13 and 29.
What is the largest prime factor of the number 600851475143 ?
Go to the thread for problem 3 in the forum.
import math
n=600851475143
i=2
print math.sqrt(n)
while i< =math.sqrt(n):
if (n%i==0):
while (n%i==0): n=n/i
print i;
i=i+1
print n
Each new term in the Fibonacci sequence is generated by adding the previous two terms. By starting with 1 and 2, the first 10 terms will be:
1, 2, 3, 5, 8, 13, 21, 34, 55, 89, ...
By considering the terms in the Fibonacci sequence whose values do not exceed four million, find the sum of the even-valued terms.
Go to the thread for problem 2 in the forum.
Even(是偶数的意思……(─.─|||
f=[1,1,1]
i=2
ans=0
while f[i]+f[i-1]<4000000:
i=i+1
f.append(f[i-1]+f[i-2])
if (f[i]%2==0): ans=ans+f[i]
print sum(f)-1,f,ans
Problem A
Chess Queen
Input: Standard Input
Output: Standard Output
王后互相攻击的前提是在一行，一列，或一对角线。:
|
|
在 (NxM) 的棋盘上 2 王后互相攻击，求方案数.
输入数据不超过 5000 行 ，每行为M and N (0< M, N£106) ,数据以0 0结尾.
每行一个整数表示方案数，保证它在u64范围内.
|
2 2 100 223 2300 1000 0 0 |
12 10907100 11514134000 |
Problemsetter: Shahriar Manzoor
Special Thanks to: Mohammad Mahmudur Rahman
首先，一个矩形的长宽若为m,n(m>=n)
那么它一个方向的对角线应为1..(n-1)各2条,n有(m-n+1)条
知道这个的化，本题就转化为，在一列一行或一对角线任取2点，有几种取法。
#include<cstdio>
#include<algorithm>
#include<functional>
#include<iostream>
#include<cstdlib>
#include<cstring>
using namespace std;
#define MAXN (1000000+10)
unsigned long long n,m;
int main()
{
while (cin>>n>>m&&n&&m)
{
if (n>m) swap(n,m);
cout<<n*m*(n+m-2)+2*n*(n-1)*(3*m-n-1)/3<<endl;
}
return 0;
}