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.

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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).

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

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Sample Input 2 2 1 100 3 2 5 3 0 1 100

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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.

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Author FZU 思路：中國剩余定理+容斥+擴展歐幾里得; 比賽時打了將近2個小時，最后還因為快速冪中的一個模沒寫而超時 GG啊。 其實題解的思路和我的思路有些不同，題解是容斥的時候將7加入一起用中國剩余定理求解，這時求出來的通解直接是7 的倍數，因為%7=0，加入求同余方程組的解； 因為只要符合模這些數中的一條就可以了，如果直接求解會重復，所以用容斥原理。加入其中求得一個通解,記為t+kmod； 然后y<=t+kmod<=x;移項可以求出k的解的個數，然后根據容斥這里是奇減偶加。 然后我的思路是沒有加入7求同余。 而是用求出那些個的數的同余方程組的解然后t+kmod=7r；再用擴展歐幾里得，求k的通解，但在這之前我先求x<=t+kmod<=y;的k的范圍。 擴歐求得的b+7p,代入前面的不等式解p的范圍，求得p有多少個，然后容斥奇減偶加p； 1 #include<stdio.h> 2 #include<algorithm> 3 #include<iostream> 4 #include<string.h> 5 #include<queue> 6 #include<stdlib.h> 7 #include<iostream> 8 #include<vector> 9 #include<map> 10 #include<set> 11 #include<math.h> 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 pair<LL,LL>P(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; i<n; i++) 46 { 47 scanf("%lld %lld",&ans[i].x,&ans[i].y); 48 } 49 LL anw=0; 50 int s; 51 for(j=1; j<(1<<n); j++) 52 { 53 int cr=0; 54 for(s=0; s<n; s++) 55 { 56 if(j&(1<<s)) 57 { 58 mm[cr++]=s; 59 } 60 } 61 LL mod=1; 62 LL acm=0; 63 for(i=0; i<cr; i++) 64 { 65 mod*=ans[mm[i]].x; 66 } 67 for(i=0; i<cr; i++) 68 { 69 LL mod1=mod/ans[mm[i]].x; 70 LL ni=quick(mod1,ans[mm[i]].x-2,ans[mm[i]].x); 71 acm=(acm+mul(mul(mod1,ni,mod),ans[mm[i]].y,mod))%mod; 72 } 73 anw=acm; 74 LL ctx=0; 75 if(anw>y) 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*nx<x) 86 { 87 nx++; 88 } if(cha<0)nx=0; 89 ny=cha1/mod; 90 { 91 pair<LL ,LL>AK=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 pair<LL,LL>P(LL n,LL m) 160 { 161 if(m==0) 162 { 163 pair<LL,LL>ak; 164 ak=make_pair(1,0); 165 return ak; 166 } 167 else 168 { 169 pair<LL,LL>A=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<stdio.h> 2 #include<algorithm> 3 #include<iostream> 4 #include<string.h> 5 #include<queue> 6 #include<stdlib.h> 7 #include<iostream> 8 #include<vector> 9 #include<map> 10 #include<set> 11 #include<math.h> 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 pair<LL,LL>P(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; i<n; i++) 45 { 46 scanf("%lld %lld",&ans[i].x,&ans[i].y); 47 } 48 LL anw=0; 49 LL s; 50 for(j=1; j<(1<<n); j++) 51 { LL mod=1; 52 int cr=0; 53 for(s=0; s<n; s++) 54 { 55 if(j&(1<<s)) 56 { 57 mm[cr++]=s; 58 mod*=ans[s].x; 59 } 60 }mod*=7; 61 LL acm=0; 62 for(i=0; i<cr; i++) 63 { 64 LL mod1=mod/ans[mm[i]].x; 65 LL ni=quick(mod1,ans[mm[i]].x-2,ans[mm[i]].x); 66 acm=(acm+mul(mul(mod1,ni,mod),ans[mm[i]].y,mod))%mod; 67 } 68 acm%=mod; 69 acm+=mod; 70 acm%=mod; 71 anw=acm; 72 LL ctx=0; 73 if(anw>y) 74 { 75 continue; 76 } 77 else 78 { 79 if(anw<x) 80 { 81 LL ax=x-anw-1; 82 LL ay=y-anw; 83 ctx+=ay/mod-ax/mod; 84 } 85 else 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 pair<LL,LL>P(LL n,LL m) 148 { 149 if(m==0) 150 { 151 pair<LL,LL>ak; 152 ak=make_pair(1,0); 153 return ak; 154 } 155 else 156 { 157 pair<LL,LL>A=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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轉載于:https://www.cnblogs.com/zzuli2sjy/p/5716795.html

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