HDU4983 Goffi and GCD （尤拉函式求解gcd個數）
阿新 • • 發佈：2019-02-20
題意簡潔明瞭，大意就是說有這樣一條公式----<span style="font-family: 'Times New Roman'; font-size: 14px;"> </span><span class="MathJax_Preview" style="color: rgb(136, 136, 136); font-family: 'Times New Roman'; font-size: 14px;"></span><span class="MathJax" id="MathJax-Element-1-Frame" role="textbox" aria-readonly="true" style="display: inline; font-size: 14px; word-wrap: normal; white-space: nowrap; float: none; direction: ltr; max-width: none; max-height: none; min-width: 0px; min-height: 0px; border: 0px; padding: 0px; margin: 0px; font-family: 'Times New Roman';"><nobr style="transition: none; border: 0px; padding: 0px; margin: 0px; max-width: none; max-height: none; min-width: 0px; min-height: 0px; vertical-align: 0px;"><span class="math" id="MathJax-Span-1" style="transition: none; display: inline-block; position: static; border: 0px; padding: 0px; margin: 0px; vertical-align: 0px; width: 15.329em;"><span style="transition: none; display: inline-block; position: relative; border: 0px; padding: 0px; margin: 0px; vertical-align: 0px; width: 14.844em; height: 0px; font-size: 14.42px;"><span style="transition: none; position: absolute; border: 0px; padding: 0px; margin: 0px; vertical-align: 0px; clip: rect(1.668em 1000em 3.193em -0.621em); top: -2.701em; left: 0.003em;"><span class="mrow" id="MathJax-Span-2" style="transition: none; display: inline; position: static; border: 0px; padding: 0px; margin: 0px; vertical-align: 0px;"><span class="mo" id="MathJax-Span-3" style="transition: none; display: inline; position: static; border: 0px; padding: 0px; margin: 0px; vertical-align: 0px; font-family: MathJax_Main;">gcd</span><span class="mo" id="MathJax-Span-4" style="transition: none; display: inline; position: static; border: 0px; padding: 0px; margin: 0px; vertical-align: 0px; font-family: MathJax_Main;">(</span><span class="mi" id="MathJax-Span-5" style="transition: none; display: inline; position: static; border: 0px; padding: 0px; margin: 0px; vertical-align: 0px; font-family: MathJax_Math-italic;">n</span><span class="mo" id="MathJax-Span-6" style="transition: none; display: inline; position: static; border: 0px; padding: 0px 0px 0px 0.212em; margin: 0px; vertical-align: 0px; font-family: MathJax_Main;">−</span><span class="mi" id="MathJax-Span-7" style="transition: none; display: inline; position: static; border: 0px; padding: 0px 0px 0px 0.212em; margin: 0px; vertical-align: 0px; font-family: MathJax_Math-italic;">a</span><span class="mo" id="MathJax-Span-8" style="transition: none; display: inline; position: static; border: 0px; padding: 0px; margin: 0px; vertical-align: 0px; font-family: MathJax_Main;">,</span><span class="mi" id="MathJax-Span-9" style="transition: none; display: inline; position: static; border: 0px; padding: 0px 0px 0px 0.142em; margin: 0px; vertical-align: 0px; font-family: MathJax_Math-italic;">n</span><span class="mo" id="MathJax-Span-10" style="transition: none; display: inline; position: static; border: 0px; padding: 0px; margin: 0px; vertical-align: 0px; font-family: MathJax_Main;">)</span><span class="mo" id="MathJax-Span-11" style="transition: none; 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display: inline; position: static; border: 0px; padding: 0px 0px 0px 0.212em; margin: 0px; vertical-align: 0px; font-family: MathJax_Math-italic;">b</span><span class="mo" id="MathJax-Span-17" style="transition: none; display: inline; position: static; border: 0px; padding: 0px; margin: 0px; vertical-align: 0px; font-family: MathJax_Main;">,</span><span class="mi" id="MathJax-Span-18" style="transition: none; display: inline; position: static; border: 0px; padding: 0px 0px 0px 0.142em; margin: 0px; vertical-align: 0px; font-family: MathJax_Math-italic;">n</span><span class="mo" id="MathJax-Span-19" style="transition: none; display: inline; position: static; border: 0px; padding: 0px; margin: 0px; vertical-align: 0px; font-family: MathJax_Main;">)</span><span class="mo" id="MathJax-Span-20" style="transition: none; display: inline; position: static; border: 0px; padding: 0px 0px 0px 0.281em; margin: 0px; vertical-align: 0px; font-family: MathJax_Main;">=</span><span class="msubsup" id="MathJax-Span-21" style="transition: none; 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display: inline; position: static; border: 0px; padding: 0px; margin: 0px; vertical-align: 0px; font-size: 12px; font-family: MathJax_Math-italic;">k</span><span style="transition: none; display: inline-block; position: static; border: 0px; padding: 0px; margin: 0px; vertical-align: 0px; width: 0px; height: 2.292em;"></span></span></span></span></span><span style="transition: none; display: inline-block; position: static; border: 0px; padding: 0px; margin: 0px; vertical-align: 0px; width: 0px; height: 2.708em;"></span></span></span><span style="transition: none; display: inline-block; position: static; border-width: 0px 0px 0px 0.004em; border-left-style: solid; border-color: initial; padding: 0px; margin: 0px; vertical-align: -0.354em; overflow: hidden; width: 0px; height: 1.289em;"></span></span></nobr></span><span style="font-family: 'Times New Roman'; font-size: 14px;">.。</span>
<span style="font-family:Times New Roman;">求在給出輸入n，k的情況下，滿足公式條件的對數，並將結果取模。此題勿忘一個條件！！！即（1，2）和（2，1）是兩對！這個坑找了我好久23333_(:з」∠)_。。。</span><span style="font-family:Times New Roman;">思路：顯然一個gcd在取最大值情況下，值為n，故k最大值為2，且k == 2時，結果只有一種，當n == 1時，結果也只有一種，故以上兩種情況直接輸出1。當k > 2時，顯然沒有滿足條件的gcd數對，故為零。此時我們只需要求k == 1的情況。</span>
<span style="font-family:Times New Roman;">設gcd（n - a，n）值為k，易知gcd(n - b，n) == n / k，從1到sqrt(n * 1.0)遍歷列舉k，此時應用尤拉函式，求出當滿足條件時候的值。設gcd（n - a，n）的個數為ans，gcd（n - b，n）為res。對於k，當k*k == n時，結果為ans * res，反之，為2 * ans * res！這就是我前面說的那個坑啊_(:з」∠)_！！！</span><span style="font-family:Times New Roman;"> </span>
<span style="font-family:Times New Roman;">程式碼：</span>#include<bits/stdc++.h>
#define mod 1000000007
using namespace std;
long long euler(long long x) //尤拉函式
{
long long res = x;
for(int i = 2;i * i <= x; i++)
{
if(x % i == 0)
{
res = res / i * (i - 1);
while(x % i == 0)
x /= i;
}
}
if(x > 1)
res = res / x * (x - 1);
return res;
}
int main()
{
long long n,k;
while(cin>>n>>k)
{
long long tot = 0;
if(k == 2 || n == 1)
{
printf("1\n");
continue;
}
if(k > 2)
{
printf("0\n");
continue;
}
for(int i = 1;i <= sqrt(n * 1.0);i++)
{
if(n % i == 0)
{
long long ans = euler(n / i);
long long res = euler(i);
if(i * i != n)
res = res % mod * 2;
tot = (tot + res * ans % mod) % mod;
}
}
printf("%lld\n",tot);
}
return 0;
}