sdutoj 2608 Alice and Bob
http://acm.sdut.edu.cn/sdutoj/problem.php?action=showproblem&problemid=2608
Alice and Bob
Time Limit: 1000ms Memory limit: 65536K 有疑问？点这里^_^
题目描述
Alice and Bob like playing games very much.Today, they introduce a new game.
There is a polynomial like this: (a0*x^(2^0)+1) * (a1 * x^(2^1)+1)*.......*(an-1 * x^(2^(n-1))+1). Then Alice ask Bob Q questions. In the expansion of the Polynomial, Given an integer P, please tell the coefficient of the x^P.
Can you help Bob answer these questions?
输入
For each case, the first line contains a number n, then n numbers a0, a1, .... an-1 followed in the next line. In the third line is a number Q, and then following Q numbers P.
1 <= T <= 20
1 <= n <= 50
0 <= ai <= 100
Q <= 1000
0 <= P <= 1234567898765432
输出
示例输入
1
2
2 1
2
3
4
示例输出
2
0
提示
来源
示例程序
分析：
给出一个多项式：(a0*x^(2^0)+1) * (a1 * x^(2^1)+1)*.......*(an-1 * x^(2^(n-1))+1)，输入P，求X^p 前边的系数。
将p转换成一个二进制的数，然后分别乘上系数。
AC代码：
1 #include<stdio.h> 2 #include<string.h> 3 int a[55],b[55]; 4 int main() 5 { 6 int t; 7 scanf("%d",&t); 8 while(t--) 9 { 10 int n,p,i,j; 11 scanf("%d",&n); 12 memset(a,0,sizeof(a)); 13 for(i=0;i<n;i++) 14 scanf("%d",&a[i]); 15 scanf("%d",&p); 16 int count,ans; 17 long long c; 18 while(p--) 19 { 20 count=1,ans=0; 21 scanf("%lld",&c); 22 if(c==0) 23 { 24 printf("1\n"); 25 continue; 26 } 27 28 while(c) 29 { 30 b[ans++]=c%2; 31 c/=2; 32 } 33 for(i=0;i<ans;i++) 34 if(b[i]) 35 { 36 count=count*a[i]%2012; 37 } 38 printf("%d\n",count); 39 } 40 } 41 return 0; 42 }
官方标程：
1 #include <iostream> 2 #include <cstdio> 3 #include <cstdlib> 4 #include <cstring> 5 #include <algorithm> 6 using namespace std; 7 const int mod = 2012; 8 int a[70], n; 9 int Q; 10 #define ll long long 11 ll w[60]; 12 #define fuck while(1) 13 int main() { 14 int t; 15 freopen("data.in", "r", stdin); 16 // freopen("data.out", "w", stdout); 17 scanf("%d", &t); 18 w[0] = 1; 19 for(int i = 1; i < 60; i++) w[i] = (w[i - 1] << 1); 20 while(t--) { 21 scanf("%d", &n); 22 if(n <= 0 || n > 50) fuck; 23 memset(a, 0, sizeof(a)); 24 for(int i = 0; i < n; i++) { 25 scanf("%d", &a[i]); 26 if(a[i] < 0 || a[i] > 100) fuck; 27 } 28 scanf("%d", &Q); 29 while(Q--) { 30 ll x; 31 scanf("%I64d", &x); 32 if(x > 1234567898765432ll) fuck; 33 int ret = 1; 34 for(int i = 0; x; i++, x /= 2) { 35 if((x & 1)) { 36 ret = ret * a[i] % mod; 37 } 38 } 39 printf("%d\n", ret); 40 } 41 } 42 return 0; 43 }
数据生成程序：
1 #include <iostream> 2 #include <cstdio> 3 #include <cstdlib> 4 #include <algorithm> 5 #include <time.h> 6 using namespace std; 7 #define ll long long 8 #define mod 1234567898765432ll 9 ll r() { 10 ll a = rand(); 11 a *= rand(); 12 a %= 100000000; 13 if(a < 0) a += 100000000; 14 return a; 15 } 16 ll maxx; 17 ll rr() { 18 int xx = rand() % 100; 19 if(xx == 0) return 0; 20 return r() * r() % maxx; 21 } 22 int main() { 23 int t = 20; 24 freopen("data.in", "w", stdout); 25 cout<<t<<endl; 26 srand(time(NULL)); 27 while(t--) { 28 int n = rand() % 50 + 1; 29 cout<<n<<endl; 30 for(int i = 0; i < n; i++) { 31 int a = rand() % 100; 32 if(i) cout<<" "; 33 cout<<a; 34 } 35 maxx = (1ll << (n)); 36 maxx = maxx + maxx / 10000; 37 cout<<endl; 38 int Q = rand() % 1000 + 1; 39 ll woca = (1ll << n); 40 Q = Q % woca + 1; 41 cout<<Q<<endl; 42 while(Q--) { 43 cout<<rr() % mod<<endl; 44 } 45 } 46 return 0; 47 }
