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http://lightoj.com/volume_showproblem.php?problem=1336
Description
Sigma function is an interesting function in Number Theory. It is denoted by the Greek letter Sigma (σ). This function actually denotes the sum of all divisors of a number. For example σ(24) = 1+2+3+4+6+8+12+24=60. Sigma of small numbers is easy to find but for large numbers it is very difficult to find in a straight forward way. But mathematicians have discovered a formula to find sigma. If the prime power decomposition of an integer is
Then we can write,
For some n the value of σ(n) is odd and for others it is even. Given a value n, you will have to find how many integers from 1 to n have even value of σ.
Input
Input starts with an integer T (≤ 100), denoting the number of test cases.
Each case starts with a line containing an integer n (1 ≤ n ≤ 1012).
Output
For each case, print the case number and the result.
Sample Input
4
3
10
100
1000
Sample Output
Case 1: 1
Case 2: 5
Case 3: 83
Case 4: 947
题目大意：给你一个数n，让你求从1到n中因子和为偶数的数共有多少个，可以用唯一分定理的公式
来找（我这样写过，不过超时）
那么我们可以通过超时的代码将100中不满足的数打出来，如下：
2 3
1 1
4 7
8 15
9 13
16 31
18 39
25 31
32 63
36 91
49 57
50 93
64 127
72 195
81 121
98 171
100 217
发现100中，因子和为奇数的有：
1 2 4 6 9 16 18 25 32 36 49 50 64 72 81 98 100
有没有发现，这些数字有一个规律，他们是 x^2, 2*x, 2*x^2, 只要100中的数满足这三个中的一个，那么，这个数就是不满足的，
总数-不满足的个数 = 满足的个数
我们还可以发现：当x为偶数时2*x和x^2会有重复的部分
当x为奇数时2*x和2*x^2会有重复的部分
那么我们可以将2*x省去，我们求求出x^2的个数和2*x^2的个数，然后用总数减去它们的个数即可
AC代码：
#include<stdio.h> #include<math.h> #include<string.h> #include<stdlib.h> #include<algorithm> using namespace std; typedef long long ll; const int N = 1e6 + 10; int main() { int t, p = 0; ll n; scanf("%d", &t); while(t--) { p++; ll x, y; scanf("%lld", &n); x = (ll)sqrt(n);//计算n中x^2的个数 y = (ll)sqrt(1.0 * n / 2);//计算n中2*x^2的个数 printf("Case %d: %lld\n", p, n - x - y); } return 0; }
TLE代码（可以用来打表找规律）：
#include<stdio.h> #include<math.h> #include<string.h> #include<stdlib.h> #include<algorithm> using namespace std; typedef long long ll; const int N = 1e6 + 10; int prime[N], k; bool Isprime[N]; void Prime() { k = 0; memset(Isprime, true, sizeof(Isprime)); Isprime[1] = false; for(int i = 2 ; i < N ; i++) { if(Isprime[i]) { prime[k++] = i; for(int j = 2 ; j * i < N ; j++) Isprime[j * i] = false; } } } ll solve(ll n) { ll x, y, a; ll num = 1; for(ll i = 0 ; i < k && prime[i] * prime[i] <= n ; i++) { x = 0; if(n % prime[i] == 0) { while(n % prime[i] == 0) { x++; n /= prime[i]; } x += 1; a = 1; for(ll j = 1 ; j <= x ; j++) a *= prime[i]; y = a - 1; num *= y /(prime[i] - 1); } } if(n > 1) num *= ((n * n - 1) / (n - 1)); return num; } int main() { Prime(); int t, x = 0; ll n; scanf("%d", &t); while(t--) { x++; scanf("%lld", &n); ll num = 0; for(ll i = 2 ; i <= n ; i++) { ll m = solve(i); if(m % 2 != 0) { printf("%lld %lld\n", i, m); num++; } } printf("Case %d: %lld\n", x, n - 1 - num); } return 0; }
LightOJ 13361336 - Sigma Function （找规律 + 唯一分解定理）
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原文地址：http://www.cnblogs.com/qq2424260747/p/4930861.html
