| 记录编号 |
158481 |
评测结果 |
AAAAAAAAAA |
| 题目名称 |
[CQOI2015]多项式 |
最终得分 |
100 |
| 用户昵称 |
 Asm.Def |
是否通过 |
通过 |
| 代码语言 |
C++ |
运行时间 |
0.065 s |
| 提交时间 |
2015-04-15 14:44:22 |
内存使用 |
0.43 MiB |
显示代码纯文本
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <ctime>
#include <cctype>
#include <algorithm>
#include <cmath>
using namespace std;
//#define USEFREAD
#ifdef USEFREAD
#define InputLen 1000000
char *ptr=(char *)malloc(InputLen);
#define getc() (*(ptr++))
#else
#define getc() (getchar())
#endif
#define SetFile(x) (freopen(#x".in", "r", stdin), freopen(#x".out", "w", stdout))
template<class T>inline void getd(T &x){
char ch = getc();bool neg = false;
while(!isdigit(ch) && ch != '-')ch = getc();
if(ch == '-')ch = getc(), neg = true;
x = ch - '0';
while(isdigit(ch = getc()))x = x * 10 - '0' + ch;
if(neg)x = -x;
}
/***********************************************************************/
typedef unsigned uint;
const uint mod = 3389, maxn = 10000, base = 10000;
uint t, dmax, An;
#include<iostream>
#include<string>
#include<iomanip>
#include<algorithm>
#define MAXN 9999
#define MAXSIZE maxn
#define DLEN 4
struct BigNum
{
int a[maxn]; //可以控制大数的位数
int len; //大数长度
BigNum(){ len = 1;memset(a,0,sizeof(a)); } //构造函数
BigNum(const int); //将一个int类型的变量转化为大数
BigNum(const char*); //将一个字符串类型的变量转化为大数
BigNum(const BigNum &); //拷贝构造函数
BigNum &operator=(const BigNum &); //重载赋值运算符，大数之间进行赋值运算
friend istream& operator>>(istream&, BigNum&); //重载输入运算符
friend ostream& operator<<(ostream&, BigNum&); //重载输出运算符
BigNum operator+(const BigNum &) const; //重载加法运算符，两个大数之间的相加运算
BigNum operator-(const BigNum &) const; //重载减法运算符，两个大数之间的相减运算
BigNum operator*(const BigNum &) const; //重载乘法运算符，两个大数之间的相乘运算
BigNum operator/(const int &) const; //重载除法运算符，大数对一个整数进行相除运算
BigNum operator^(const int &) const; //大数的n次方运算
int operator%(const int &) const; //大数对一个int类型的变量进行取模运算
bool operator>(const BigNum & T)const; //大数和另一个大数的大小比较
bool operator>(const int & t)const; //大数和一个int类型的变量的大小比较
void print(); //输出大数
}M(0), N(0), fd = 1;
BigNum::BigNum(const int b) //将一个int类型的变量转化为大数
{
int c,d = b;
len = 0;
memset(a,0,sizeof(a));
while(d > MAXN)
{
c = d - (d / (MAXN + 1)) * (MAXN + 1);
d = d / (MAXN + 1);
a[len++] = c;
}
a[len++] = d;
}
BigNum::BigNum(const char*s) //将一个字符串类型的变量转化为大数
{
int t,k,index,l,i;
memset(a,0,sizeof(a));
l=strlen(s);
len=l/DLEN;
if(l%DLEN)
len++;
index=0;
for(i=l-1;i>=0;i-=DLEN)
{
t=0;
k=i-DLEN+1;
if(k<0)
k=0;
for(int j=k;j<=i;j++)
t=t*10+s[j]-'0';
a[index++]=t;
}
}
BigNum::BigNum(const BigNum & T) : len(T.len) //拷贝构造函数
{
int i;
memset(a,0,sizeof(a));
for(i = 0 ; i < len ; i++)
a[i] = T.a[i];
}
BigNum & BigNum::operator=(const BigNum & n) //重载赋值运算符，大数之间进行赋值运算
{
int i;
len = n.len;
memset(a,0,sizeof(a));
for(i = 0 ; i < len ; i++)
a[i] = n.a[i];
return *this;
}
istream& operator>>(istream & in, BigNum & b) //重载输入运算符
{
char ch[MAXSIZE*4];
int i = -1;
in>>ch;
int l=strlen(ch);
int count=0,sum=0;
for(i=l-1;i>=0;)
{
sum = 0;
int t=1;
for(int j=0;j<4&&i>=0;j++,i--,t*=10)
{
sum+=(ch[i]-'0')*t;
}
b.a[count]=sum;
count++;
}
b.len =count++;
return in;
}
ostream& operator<<(ostream& out, BigNum& b) //重载输出运算符
{
int i;
cout << b.a[b.len - 1];
for(i = b.len - 2 ; i >= 0 ; i--)
{
cout.width(DLEN);
cout.fill('0');
cout << b.a[i];
}
return out;
}
BigNum BigNum::operator+(const BigNum & T) const //两个大数之间的相加运算
{
BigNum t(*this);
int i,big; //位数
big = T.len > len ? T.len : len;
for(i = 0 ; i < big ; i++)
{
t.a[i] +=T.a[i];
if(t.a[i] > MAXN)
{
t.a[i + 1]++;
t.a[i] -=MAXN+1;
}
}
if(t.a[big] != 0)
t.len = big + 1;
else
t.len = big;
return t;
}
BigNum BigNum::operator-(const BigNum & T) const //两个大数之间的相减运算
{
int i,j,big;
bool flag;
BigNum t1,t2;
if(*this>T)
{
t1=*this;
t2=T;
flag=0;
}
else
{
t1=T;
t2=*this;
flag=1;
}
big=t1.len;
for(i = 0 ; i < big ; i++)
{
if(t1.a[i] < t2.a[i])
{
j = i + 1;
while(t1.a[j] == 0)
j++;
t1.a[j--]--;
while(j > i)
t1.a[j--] += MAXN;
t1.a[i] += MAXN + 1 - t2.a[i];
}
else
t1.a[i] -= t2.a[i];
}
t1.len = big;
while(t1.a[len - 1] == 0 && t1.len > 1)
{
t1.len--;
big--;
}
if(flag)
t1.a[big-1]=0-t1.a[big-1];
return t1;
}
BigNum BigNum::operator*(const BigNum & T) const //两个大数之间的相乘运算
{
BigNum ret;
int i,j,up;
int temp,temp1;
for(i = 0 ; i < len ; i++)
{
up = 0;
for(j = 0 ; j < T.len ; j++)
{
temp = a[i] * T.a[j] + ret.a[i + j] + up;
if(temp > MAXN)
{
temp1 = temp - temp / (MAXN + 1) * (MAXN + 1);
up = temp / (MAXN + 1);
ret.a[i + j] = temp1;
}
else
{
up = 0;
ret.a[i + j] = temp;
}
}
if(up != 0)
ret.a[i + j] = up;
}
ret.len = i + j;
while(ret.a[ret.len - 1] == 0 && ret.len > 1)
ret.len--;
return ret;
}
BigNum BigNum::operator/(const int & b) const //大数对一个整数进行相除运算
{
BigNum ret;
int i,down = 0;
for(i = len - 1 ; i >= 0 ; i--)
{
ret.a[i] = (a[i] + down * (MAXN + 1)) / b;
down = a[i] + down * (MAXN + 1) - ret.a[i] * b;
}
ret.len = len;
while(ret.a[ret.len - 1] == 0 && ret.len > 1)
ret.len--;
return ret;
}
int BigNum::operator %(const int & b) const //大数对一个int类型的变量进行取模运算
{
int i,d=0;
for (i = len-1; i>=0; i--)
{
d = ((d * (MAXN+1))% b + a[i])% b;
}
return d;
}
BigNum BigNum::operator^(const int & n) const //大数的n次方运算
{
BigNum t,ret(1);
int i;
if(n<0)
exit(-1);
if(n==0)
return 1;
if(n==1)
return *this;
int m=n;
while(m>1)
{
t=*this;
for( i=1;i<<1<=m;i<<=1)
{
t=t*t;
}
m-=i;
ret=ret*t;
if(m==1)
ret=ret*(*this);
}
return ret;
}
bool BigNum::operator>(const BigNum & T) const //大数和另一个大数的大小比较
{
int ln;
if(len > T.len)
return true;
else if(len == T.len)
{
ln = len - 1;
while(a[ln] == T.a[ln] && ln >= 0)
ln--;
if(ln >= 0 && a[ln] > T.a[ln])
return true;
else
return false;
}
else
return false;
}
bool BigNum::operator >(const int & t) const //大数和一个int类型的变量的大小比较
{
BigNum b(t);
return *this>b;
}
void BigNum::print() //输出大数
{
int i;
cout << a[len - 1];
for(i = len - 2 ; i >= 0 ; i--)
{
cout.width(DLEN);
cout.fill('0');
cout << a[i];
}
cout << endl;
}
struct Mat{uint a, b, c, d;Mat(uint w,uint x,uint y,uint z):a(w),b(x),c(y),d(z){}}F(1234,0,2289,1);
inline void operator *= (Mat &x, const Mat &y){
uint a = (x.a * y.a + x.b * y.c) % mod,
b = (x.a * y.b + x.b * y.d) % mod,
c = (x.c * y.a + x.d * y.c) % mod,
d = (x.c * y.b + x.d * y.d) % mod;
x = Mat(a, b, c, d);
}
inline Mat powmod(Mat a, uint n){
Mat ans(1,0,0,1);
while(n){
if(n & 1)ans *= a;
a *= a;
n >>= 1;
}
return ans;
}
inline Mat powmod(Mat a, const BigNum &n){
const int *it = n.a, *end = it + n.len;
Mat ans(1,0,0,1);
while(it < end){
if(*it)ans *= powmod(a, *it);
a = powmod(a, base);
++it;
}
return ans;
}
inline void init(){
cin >> N;getd(t);cin >> M;
int tmp = *N.a - *M.a;if(tmp < 0)tmp += base;
dmax = tmp;
Mat tf = powmod(F, M);
An = (tf.a + tf.c) % mod;
}
inline void work(){
BigNum Ans = An;
uint d;
for(d = 1;d <= dmax;++d){
fd = (fd * (M + d) * t) / d;
An = (An * 1234 + 2289) % mod;
Ans = Ans + fd * An;
}
Ans.print();
}
int main(){
#ifdef DEBUG
freopen("test.txt", "r", stdin);
#else
SetFile(cqoi15_polynomial);
#endif
#ifdef USEFREAD
fread(ptr,1,InputLen,stdin);
#endif
init();
work();
#ifdef DEBUG
printf("\n%.3lf sec \n", (double)clock() / CLOCKS_PER_SEC);
#endif
return 0;
}