[hdu6434]Problem I. Count
Posted memory-of-winter
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题目大意:$T(Tleqslant 10^5)$组数据,每组数据给你$n(nleqslant 2 imes 10^7)$,求$sumlimits_{i=1}^nsumlimits_{j=1}^{i-1}[(i+j,i-j)==1]$
题解:
$$
defdsum{displaystylesumlimits}
egin{align*}
&dsum_{i=1}^ndsum_{j=1}^{i-1}[(i+j,i-j)=1]\
&令k=i-j\
=&dsum_{i=1}^ndsum_{k=1}^{i-1}[(2i-k,k)=1]\
=&dsum_{i=1}^ndsum_{k=1}^{i-1}[(2i,k)=1]\
end{align*}
$$
$$
herefore
(2i,k)=1Rightarrow
egin{cases}
(i,k)=1\
(2,k)=1\
end{cases}\
当i为偶数时:\
(i,k)=1\
Rightarrow (2,k)=1\
herefore ans=varphi(i)\
当i为奇数时:\
(2,k)=1\
Rightarrow k为奇数\
herefore k必为与i互质的数的奇数\
ecause (i,k)=1Rightarrow(i,i-k)=1\
herefore 当i为奇数的时候,k奇偶各半\
herefore ans=dfrac{varphi(i)}2
$$
卡点:无
C++ Code:
#include <cstdio> #define maxn 20000010 int Tim, n; long long pre[maxn]; int plist[maxn << 3], pt, phi[maxn]; bool isp[maxn]; void sieve(int n) { phi[1] = 1; pre[1] = 0; isp[1] = true; for (int i = 2; i < n; i++) { if (!isp[i]) { plist[pt++] = i; phi[i] = i - 1; } for (int j = 0; j < pt && i * plist[j] < n; j++) { int tmp = i * plist[j]; isp[tmp] = true; if (i % plist[j] == 0) { phi[tmp] = phi[i] * plist[j]; break; } phi[tmp] = phi[i] * phi[plist[j]]; } pre[i] = pre[i - 1] + ((i & 1) ? phi[i] / 1 : phi[i]); } } int main() { sieve(maxn); scanf("%d", &Tim); while (Tim --> 0) { scanf("%d", &n); printf("%lld ", pre[n]); } return 0; }
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