Problem I. Count

Time Limit: 4000/2000 MS (Java/Others)????Memory Limit: 524288/524288 K (Java/Others)
Total Submission(s): 42????Accepted Submission(s): 16

Problem Description Multiple query, for each n, you need to get
n i-1
∑ ∑ [gcd(i + j, i - j) = 1]
i=1 j=1

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Input On the first line, there is a positive integer T, which describe the number of queries. Next there are T lines, each line give a positive integer n, as mentioned above.
T<=1e5, n<=2e7

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Output Your output should include T lines, for each line, output the answer for the corre- sponding n.

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Sample Input 4 978 438 233 666

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Sample Output 194041 38951 11065 89963

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Source 2018 Multi-University Training Contest 10

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Recommend chendu???|???We have carefully selected several similar problems for you:??6437?6436?6435?6434?6433? 題意：求下面這個式子的值 n i-1
∑ ∑ [gcd(i + j, i - j) = 1]
i=1 j=1 分析：按照上面那個式子直接暴力打表容易得到上面式子的含義是1-n以內偶數的歐拉函數值加上奇數的歐拉函數值除以2的和（注意要快速打表，運用素數快速打表） AC代碼： #include <map> #include <set> #include <stack> #include <cmath> #include <queue> #include <cstdio> #include <vector> #include <string> #include <bitset> #include <cstring> #include <iomanip> #include <iostream> #include <algorithm> #define ls (r<<1) #define rs (r<<1|1) #define debug(a) cout << #a << " " << a << endl using namespace std; typedef long long ll; const ll maxn = 2*1e7+10; const ll mod = 998244353; const double pi = acos(-1.0); const double eps = 1e-8; ll num[maxn], sum[maxn], prim[maxn]; int main() {ios::sync_with_stdio(0);ll T, n;memset(num, 0, sizeof num);num[1] = 1;ll id = 0;for( ll i = 2; i <= maxn-10; i ++ ) {if(!num[i]) {num[i] = i - 1; prim[id++] = i;}for( ll j = 0; j < id && prim[j]*i <= maxn-10; j ++ ) {if(i % prim[j]) {num[i*prim[j]] = num[i] * (prim[j]-1);}else {num[i*prim[j]] = num[i] * prim[j];break;}}}num[1] = 0;sum[1] = 0;for( ll i = 2; i <= maxn-10; i ++ ) {if( i%2 == 0 ) {sum[i] = sum[i-1] + num[i];} else {sum[i] = sum[i-1] + num[i]/2;}}scanf("%lld",&T);while( T -- ) {scanf("%lld",&n);printf("%lld\n",sum[n]);}return 0; }

　　

轉載于:https://www.cnblogs.com/l609929321/p/9519403.html

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