//hdu 1695
GCD
Time Limit: 6000/3000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 15816 Accepted Submission(s): 6095
Problem Description
Given 5 integers: a, b, c, d, k, you're to find x in a...b, y in c...d that GCD(x, y) = k. GCD(x, y) means the greatest common divisor of x and y. Since the number of choices may be very large, you're only required to output the total number of different number pairs.
Please notice that, (x=5, y=7) and (x=7, y=5) are considered to be the same.
Yoiu can assume that a = c = 1 in all test cases.
Please notice that, (x=5, y=7) and (x=7, y=5) are considered to be the same.
Yoiu can assume that a = c = 1 in all test cases.
Input
The input consists of several test cases. The first line of the input is the number of the cases. There are no more than 3,000 cases.
Each case contains five integers: a, b, c, d, k, 0 < a <= b <= 100,000, 0 < c <= d <= 100,000, 0 <= k <= 100,000, as described above.
Each case contains five integers: a, b, c, d, k, 0 < a <= b <= 100,000, 0 < c <= d <= 100,000, 0 <= k <= 100,000, as described above.
Output
For each test case, print the number of choices. Use the format in the example.
Sample Input
2
1 3 1 5 1
1 11014 1 14409 9
Sample Output
Case 1: 9
Case 2: 736427
Hint
For the first sample input, all the 9 pairs of numbers are (1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (2, 3), (2, 5), (3, 4), (3, 5).Source
如果 ：
那么 ：
如果 ：
那么 ：
d|n == n%d=0
1 #include<cstdio> 2 #include<iostream> 3 #include<algorithm> 4 #include<cstring> 5 #include<cmath> 6 #include<vector> 7 #define N 100009 8 using namespace std; 9 #define ll long long 10 #define mem(a,b) memset(a,b,sizeof(a)) 11 int a,b,c,d,k,t; 12 int u[N],pre[N]; 13 bool vis[N]; 14 /* 15 其中μ()函数是莫比乌斯函数，定义是： 16 如果d=1 ， μ(d)=1 17 如果d为互异质数p1,p2…pk的乘积，则μ(d)=(-1)^k 18 否则，μ(d)=0 19 */ 20 void init(){ 21 u[1]=1; 22 int i,j,pree=0;//局部变量要初始化 23 for(i=2;i<=N;i++){ 24 if(!vis[i]){ 25 pre[++pree]=i; 26 u[i]=-1; 27 } 28 for(j=1;j<=pree&&i*pre[j]<=N;j++){ 29 vis[i*pre[j]]=1; 30 if(i%pre[j]==0){ 31 u[i*pre[j]]=0;//不是由多个质数组成 32 break; 33 } 34 else{ 35 u[i*pre[j]]=-u[i]; 36 } 37 } 38 } 39 } 40 int main() 41 { 42 init(); 43 scanf("%d",&t); 44 for(int i=1;i<=t;i++){ 45 scanf("%d%d%d%d%d",&a,&b,&c,&d,&k); 46 if(!k) { 47 printf("Case %d: 0 ",i);//注意输出格式 48 continue; 49 } 50 b/=k,d/=k;//gcd(kx,ky)==k 互相推出 gcd(x,y)==1 51 if(b>d) swap(b,d);//b:小的 52 ll ans1=0,ans2=0; 53 for(int i=1;i<=b;i++) ans1+=(ll)u[i]*(b/i)*(d/i);//三个因子相乘的结果可能超int 54 for(int i=1;i<=b;i++) ans2+=(ll)u[i]*(b/i)*(b/i); 55 ans1-=ans2/2;//去重 56 printf("Case %d: %lld ",i,ans1); 57 } 58 return 0; 59 }
51 NOD
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关注有一个M * N的表格，行与列分别是1 - M和1 - N，格子中间写着行与列的最大公约数Gcd(i, j)(1 <= i <= M, 1 <= j <= N)。
例如：M = 5, n = 4。
1 2 3 4 5
1 1 1 1 1 1
2 1 2 1 2 1
3 1 1 3 1 1
4 1 2 1 4 1
给出M和N，求这张表中有多少个质数。
Input
第1行：一个数T，表示后面用作输入测试的数的数量。（1 <= T <= 1000) 第2 - T + 1行：每行2个数M,N，中间用空格分隔，表示表格的宽和高。(1 <= M, N <= 5 * 10^6)
Output
共T行，每行1个数，表示表格中质数的数量。
Input示例
2 10 10 100 100
Output示例
30 2791
1 #include<cstdio> 2 #include<iostream> 3 #include<algorithm> 4 #include<cstring> 5 #include<cmath> 6 #include<vector> 7 #define N 5000009 8 using namespace std; 9 #define ll long long 10 #define mem(a,b) memset(a,b,sizeof(a)) 11 int t,n,m,last; 12 int pre[N],u[N]; 13 int sum[N],summ[N];//ll会MET 14 ll ans; 15 bool vis[N]; 16 void init(){ 17 u[1]=1; 18 int i,j,pree=0; 19 for(i=2;i<=N;i++){ 20 if(!vis[i]){ 21 pre[++pree]=i; 22 u[i]=-1; 23 } 24 for(j=1;j<=pree&&i*pre[j]<=N;j++) 25 { 26 vis[i*pre[j]]=1; 27 if(i%pre[j]==0){ 28 u[i*pre[j]]=0; 29 break; 30 } 31 else{ 32 u[i*pre[j]]=-u[i]; 33 } 34 } 35 } 36 for(i=1;i<=pree;i++){ 37 for(j=pre[i];j<=N;j+=pre[i]){ 38 sum[j]+=u[j/pre[i]]; 39 } 40 } 41 for(i=1;i<=N;i++){ 42 summ[i]=summ[i-1]+sum[i]; 43 } 44 printf("%d ",pree); 45 } 46 int main() 47 { 48 init(); 49 scanf("%d",&t); 50 while(t--) 51 { 52 scanf("%d%d",&n,&m); 53 int lim=min(n,m); 54 ans=0; 55 for(int i=1;i<=lim;i=last+1){ 56 last=min(n/(n/i),m/(m/i)); 57 ans+=(ll)(n/i)*(m/i)*(summ[last]-summ[i-1]);//i~last 的F（d)都等于 (n/i)*(m/i)
如 ：
10/7 * 12/7
10/8 * 12/8
10/9 * 12/9
10/10 * 12/10 是相等的 58 } 59 printf("%lld ",ans); 60 } 61 return 0; 62 }