1
#define _CRT_SECURE_NO_WARNINGS 1//jerry99的数列
2 #include<bits/stdc++.h>
3 int prime[40000] = { 0 };
4 int pri[6000] = { 0 };
5 int nums[3000] = { 0 };
6 int mem[3000] = { 0 };
7 int jiecheng[40000*1000] = { 0 };
8 void pre() {//标记质数1~50000
9 prime[0] = prime[1] = 1;
10 for (int i = 2; i < 40000; i++) {
11 for (int j = 2 * i; j < 40000; j += i) {
12 prime[j] = 1;
13 }
14 }
15 int k = 0;
16 for (int i = 0; i < 40000; i++) {
17 if (!prime[i])
18 pri[k++] = i;//存放在num数组里备用
19 }
20 }
21
22
23 int div2(int n) {
24 memset(mem, 0, 6000);
25 int k = 0, d, num = 0, origin = n;
26 for (int i = 0; n != 1; i++) {
27 d = pri[i];
28 while (n % d == 0) {
29 mem[k]++;
30 n /= d;
31 }
32 k++;
33 if (k >= 5133&&n>50000) return 1;
34 }
35 return 0;
36 }
37 int main() {
38 pre();//筛出质数
39 //提前分解好1~20000阶乘质因数
40
41
42 int t, n, m;
43 for (n = 1; n < 40000; n++) {
44 int k = 0, d, num = 0, origin = n;
45 for (int i = 0; pri[i]<=n; i++) {
46 d = pri[i];
47 while (origin != 1&&origin%d==0) {
48 nums[k]++;
49 origin /= d;
50 }
51 k++;
52 }
53 for (int p = 0; p < k; p++) {
54 *(jiecheng+1000*n+p) += (*(jiecheng+1000*(n-1)+p)+ nums[p]);//递推地计算质因数分解结果
55 //printf("%d ", jiecheng[n][p]);//数组一维化
56 }//printf("\n");
57 memset(nums, 0, 3000*4);
58 }
59 for (int i = 1; i < 10; i++) {
60 for (int j = 0; j < 10; j++) {
61 printf("%d ", jiecheng[i][j]);
62 }printf("\n");
63 }
64 scanf("%d", &t);
65 int judge;
66 while (t > 0) {
67 memset(mem, 0, 3000*4);
68 scanf("%d%d", &n, &m);//n分母,m分子
69 judge = div2(n);//把分母质因数分解
70
71 if (judge&&n>40000) {//如果n分解后的质因数是个超级大(>50000)的质数
72 if (m < n) printf("0\n");
73 else {
74 printf("%d\n", m - n + 1);
75 }
76 }
77 else {//小于50000的质数或者不是质数的10^9以内的
78 int cnt, j = 0;
79 for (cnt = 1; cnt < n;) {
80 //printf("%d ", *(jiecheng + 1000 * cnt + j));
81 while (mem[j] > * (jiecheng + 1000 * cnt + j)) {
82 cnt++;//每一位幂次进行比较
83 }
84
85 //printf("%d ", *(jiecheng + 1000 * cnt + j));
86 if (mem[j] <= *(jiecheng + 1000 * cnt + j)) {
87 j++;//比较完一个每次进入下一个幂次
88 }
89 if (!mem[j] && j == n) {
90 break;
91 }
92
93
94 }
95 /*for (int i = 1; i < n; i++) {
96 for (int j =0 ; j < 10; j++) {
97 printf("%d ", *(jiecheng +1000 * i + j));
98 }
99 }*/
100 //printf("比%d大的最小阶乘是 %d !\n", n, cnt);
101 if (cnt > m + 1) printf("0\n");
102 else printf("%d\n", m - cnt + 1);
103 t--;
104 }
105 }
106 return 0;
107 }
108
