V2AS问路
Lucky7(hdu5768)
阅读原文时间：2023年07月08日阅读：2

Lucky7

**Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 328    Accepted Submission(s): 130
**

Problem Description

When ?? was born, seven crows flew in and stopped beside him. In its childhood, ?? had been unfortunately fall into the sea. While it was dying, seven dolphins arched its body and sent it back to the shore. It is said that ?? used to surrounded by 7 candles when he faced a extremely difficult problem, and always solve it in seven minutes. 
?? once wrote an autobiography, which mentioned something about himself. In his book, it said seven is his favorite number and he thinks that a number can be divisible by seven can bring him good luck. On the other hand, ?? abhors some other prime numbers and thinks a number x divided by pi which is one of these prime numbers with a given remainder ai will bring him bad luck. In this case, many of his lucky numbers are sullied because they can be divisible by 7 and also has a remainder of ai when it is divided by the prime number pi.
Now give you a pair of x and y, and N pairs of ai and pi, please find out how many numbers between x and y can bring ?? good luck.

Input

On the first line there is an integer T(T≤20) representing the number of test cases.
Each test case starts with three integers three intergers n, x, y(0<=n<=15,0<x<y<1018) on a line where n is the number of pirmes. 
Following on n lines each contains two integers pi, ai where pi is the pirme and ?? abhors the numbers have a remainder of ai when they are divided by pi. 
It is guranteed that all the pi are distinct and pi!=7. 
It is also guaranteed that p1*p2*…*pn<=1018 and 0<ai<pi<=105for every i∈(1…n).

Output

For each test case, first output "Case #x: ",x=1,2,3…., then output the correct answer on a line.

Sample Input

2
2 1 100
3 2
5 3
0 1 100

Sample Output

Case #1: 7
Case #2: 14

Hint

For Case 1: 7,21,42,49,70,84,91 are the seven numbers.
For Case2: 7,14,21,28,35,42,49,56,63,70,77,84,91,98 are the fourteen numbers.

Author

FZU

思路：中国剩余定理+容斥+扩展欧几里得;

比赛时打了将近2个小时，最后还因为快速幂中的一个模没写而超时 GG啊。

其实题解的思路和我的思路有些不同，题解是容斥的时候将7加入一起用中国剩余定理求解，这时求出来的通解直接是7 的倍数，因为%7=0，加入求同余方程组的解；

因为只要符合模这些数中的一条就可以了，如果直接求解会重复，所以用容斥原理。加入其中求得一个通解,记为t+kmod；

然后y<=t+kmod<=x;移项可以求出k的解的个数，然后根据容斥这里是奇减偶加。

然后我的思路是没有加入7求同余。

而是用求出那些个的数的同余方程组的解然后t+kmod=7r；再用扩展欧几里得，求k的通解，但在这之前我先求x<=t+kmod<=y;的k的范围。

扩欧求得的b+7p,代入前面的不等式解p的范围，求得p有多少个，然后容斥奇减偶加p；

1 #include
2 #include
3 #include
4 #include
5 #include
6 #include
7 #include
8 #include
9 #include
10 #include
11 #include
12 using namespace std;
13 typedef long long LL;
14 typedef struct pp
15 {
16 LL x;
17 LL y;
18 } ss;
19 ss ans[20];
20 LL quick(LL n,LL m,LL mod);
21 LL mul(LL n, LL m,LL p);
22 pairP(LL n,LL m);
23 LL gcd(LL n, LL m);
24 LL mm[20];
25 int main(void)
26 {
27 int i,j,k;
28 scanf("%d",&k);
29 int __ca=0;
30 LL n,x,y;
31 while(k--)
32 {
33 __ca++;
34 scanf("%lld %lld %lld",&n,&x,&y);
35 LL sum;
36 sum=y/7-(x-1)/7;
37 printf("Case #%d: ",__ca);
38 if(n==0)
39 {
40 printf("%lld\n",sum);
41 }
42 else
43 {
44 LL mod=1;
45 for(i=0; iy)
76 {
77 continue;
78 }
79 else
80 {
81 LL cha=x-anw;
82 LL nx,ny;
83 LL cha1=y-anw;
84 nx=cha/mod;
85 while(anw+mod*nxAK=P(mod,7);
92 LL nxx=(AK.first*acm%7+7)%7;
93 nxx=((-nxx)%7+7)%7;
94 if(ny>=nxx)
95 {
96 LL cx=max(nx-nxx,(LL)0);
97 LL cy=ny-nxx;
98 if(cx==0)ctx=cy/7+1;else {ctx=cy/7-(cx-1)/7;}
99 }
100 }
101 }
102 if(cr%2)
103 {
104 sum-=ctx;
105 }
106 else sum+=ctx;
107 } printf("%lld\n",sum);
108 }
109 }
110 return 0;
111 }
112 LL gcd(LL n, LL m)
113 {
114 if(m==0)
115 {
116 return n;
117 }
118 else if(n%m==0)
119 {
120 return m;
121 }
122 else
123 {
124 return gcd(m,n%m);
125 }
126 }
127 LL quick(LL n,LL m,LL mod)
128 {
129 LL cnt=1;n%=mod;
130 while(m)
131 {
132 if(m&1)
133 {
134 cnt=cnt*n%mod;
135 }
136 n=n*n%mod;
137 m/=2;
138 }
139 return cnt;
140 }
141 LL mul(LL n, LL m,LL p)
142 {
143 n%=p;
144 m%=p;
145 LL ret=0;
146 while(m)
147 {
148 if(m&1)
149 {
150 ret=ret+n;
151 ret%=p;
152 }
153 m>>=1;
154 n<<=1; 155 n%=p; 156 } 157 return ret; 158 } 159 pairP(LL n,LL m)
160 {
161 if(m==0)
162 {
163 pairak;
164 ak=make_pair(1,0);
165 return ak;
166 }
167 else
168 {
169 pairA=P(m,n%m);
170 LL nx=A.second;
171 LL ny=A.first;
172 ny=ny-(n/m)*nx;
173 A.first=nx;
174 A.second=ny;
175 return A;
176 }
177 }

1 #include
2 #include
3 #include
4 #include
5 #include
6 #include
7 #include
8 #include
9 #include
10 #include
11 #include
12 using namespace std;
13 typedef long long LL;
14 typedef struct pp
15 {
16 LL x;
17 LL y;
18 } ss;
19 ss ans[20];
20 LL quick(LL n,LL m,LL mod);
21 LL mul(LL n, LL m,LL p);
22 pairP(LL n,LL m);
23 LL gcd(LL n, LL m);
24 LL mm[20];
25 int main(void)
26 {
27 int i,j,k;
28 scanf("%d",&k);
29 int __ca=0;
30 LL n,x,y;
31 while(k--)
32 {
33 __ca++;
34 scanf("%lld %lld %lld",&n,&x,&y);
35 LL sum;
36 sum=y/7-(x-1)/7;
37 if(n==0)
38 {
39 printf("Case #%d: %lld\n",__ca,sum);
40 }
41 else
42 {
43 LL mod=1;
44 for(i=0; iy)
74 {
75 continue;
76 }
77 else
78 {
79 if(anw=x)
86 { LL ay=y-anw;
87 ctx+=ay/mod+1;
88 }
89 }
90 if(cr%2)
91 {
92 sum-=ctx;
93 }
94 else sum+=ctx;
95 } printf("Case #%d: %lld\n",__ca,sum);
96 }
97 }
98 return 0;
99 }
100 LL gcd(LL n, LL m)
101 {
102 if(m==0)
103 {
104 return n;
105 }
106 else if(n%m==0)
107 {
108 return m;
109 }
110 else
111 {
112 return gcd(m,n%m);
113 }
114 }
115 LL quick(LL n,LL m,LL mod)
116 {
117 LL cnt=1;n%=mod;
118 while(m>0)
119 {
120 if(m&1)
121 {
122 cnt=cnt*n%mod;
123 }
124 n=n*n%mod;
125 m/=2;
126 }
127 return cnt;
128 }
129 LL mul(LL n, LL m,LL p)
130 {
131 n%=p;
132 m%=p;
133 LL ret=0;
134 while(m)
135 {
136 if(m&1)
137 {
138 ret=ret+n;
139 ret%=p;
140 }
141 m>>=1;
142 n<<=1; 143 n%=p; 144 } 145 return ret; 146 } 147 pairP(LL n,LL m)
148 {
149 if(m==0)
150 {
151 pairak;
152 ak=make_pair(1,0);
153 return ak;
154 }
155 else
156 {
157 pairA=P(m,n%m);
158 LL nx=A.second;
159 LL ny=A.first;
160 ny=ny-(n/m)*nx;
161 A.first=nx;
162 A.second=ny;
163 return A;
164 }
165 }

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