hdu 4927 Series 1（高精度） 2014多校训练第6场
Series 1
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 262144/262144 K (Java/Others)
Problem Description
Let A be an integral series {A
1, A
2, . . . , A
n}.
The zero-order series of A is A itself.
The first-order series of A is {B 1, B 2, . . . , B n-1},where B i = A i+1 - A i.
The ith-order series of A is the first-order series of its (i - 1)th-order series (2<=i<=n - 1).
Obviously, the (n - 1)th-order series of A is a single integer. Given A, figure out that integer.
The zero-order series of A is A itself.
The first-order series of A is {B 1, B 2, . . . , B n-1},where B i = A i+1 - A i.
The ith-order series of A is the first-order series of its (i - 1)th-order series (2<=i<=n - 1).
Obviously, the (n - 1)th-order series of A is a single integer. Given A, figure out that integer.
Input
The input consists of several test cases. The first line of input gives the number of test cases T (T<=10).
For each test case:
The first line contains a single integer n(1<=n<=3000), which denotes the length of series A.
The second line consists of n integers, describing A 1, A 2, . . . , A n. (0<=A i<=10 5)
For each test case:
The first line contains a single integer n(1<=n<=3000), which denotes the length of series A.
The second line consists of n integers, describing A 1, A 2, . . . , A n. (0<=A i<=10 5)
Output
For each test case, output the required integer in a line.
Sample Input
2 3 1 2 3 4 1 5 7 2
Sample Output
0 -5
题意：给出n个数，执行n-1次合并，每次合并时用后一个数减去前一个数，问最后剩下的那个数是多少。
分析：分析以后不难发现，最后的结果是C(n-1,0)*a[n] －C(n-1, 1) * a[n-1] ……C(n-1,n-1)*a[1]。符号位一正一负交替。主要是大数运算。
比赛的时候用C++中的string写的大数，调了几个小时，一直TLE，直到最后也没有搞出来。而队友用JAVA很快水掉了，
。比赛完搜题解，发现只有极少数人用C++写的，看来JAVA写大数确实很方便。
。比赛完搜题解，发现只有极少数人用C++写的，看来JAVA写大数确实很方便。
#include<cstdio>
#include<cstring>
using namespace std;
#define MAXN 9999
#define MAXSIZE 10
#define DLEN 4
class BigInt
{
private:
int a[500];
int len;
public:
BigInt() {len = 1; memset(a, 0, sizeof(a));}
BigInt(const int);
BigInt(const char *);
BigInt(const BigInt &);
BigInt &operator = (const BigInt &);
BigInt operator + (const BigInt &) const;
BigInt operator - (const BigInt &) const;
BigInt operator * (const BigInt &) const;
BigInt operator / (const int &) const;
bool operator > (const BigInt &T) const;
bool operator < (const BigInt &T) const;
bool operator == (const BigInt &T) const;
bool operator > (const int &t) const;
bool operator < (const int &t) const;
bool operator == (const int &t) const;
void print();
};
bool BigInt::operator == (const BigInt &T) const {
return !(*this > T) && !(T > *this);
}
bool BigInt::operator == (const int &t) const {
BigInt T = BigInt(t);
return *this == T;
}
bool BigInt::operator < (const BigInt &T) const {
return T > *this;
}
bool BigInt::operator < (const int &t) const {
return BigInt(t) > *this;
}
BigInt::BigInt(const int b) {
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;
}
BigInt::BigInt(const char *s) {
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;
}
}
BigInt::BigInt(const BigInt &T) : len(T.len)
{
int i;
memset(a, 0, sizeof(a));
for(i = 0; i < len; i++)
a[i] = T.a[i];
}
BigInt & BigInt::operator = (const BigInt &n)
{
int i;
len = n.len;
memset(a, 0, sizeof(a));
for(i = 0; i < len; i++)
a[i] = n.a[i];
return *this;
}
BigInt BigInt::operator + (const BigInt &T) const{
BigInt t(*this);
int i, big;
big = T.len > len ? T.len : len;
for(int 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;
}
BigInt BigInt::operator - (const BigInt &T) const {
int i, j, big;
bool flag;
BigInt 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[t1.len - 1] == 0 && t1.len > 1)
{
t1.len--;
big--;
}
if(flag)
t1.a[big-1] = 0 - t1.a[big-1];
return t1;
}
BigInt BigInt::operator * (const BigInt &T) const {
BigInt ret;
int i, j, up;
int tmp, temp;
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)
{
tmp = temp - temp / (MAXN + 1) * (MAXN + 1);
up = temp / (MAXN + 1);
ret.a[i + j] = tmp;
}
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;
}
BigInt BigInt::operator / (const int &b) const {
BigInt 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;
}
bool BigInt::operator > (const BigInt &T) const {
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 BigInt::operator > (const int &t) const {
BigInt b(t);
return *this > b;
}
void BigInt::print()
{
printf("%d",a[len-1]);
for(int i = len - 2; i >= 0; i--)
printf("%04d", a[i]);
printf("\n");
}
int main()
{
int n, T, i, j, a[3005];
scanf("%d",&T);
while(T--)
{
scanf("%d",&n);
for(i = 1; i <= n; i++)
scanf("%d",&a[i]);
BigInt positive(0), negative(0), C(1), tmp;
for(i = 0; i <= n - 1; i++)
{
if(i > 0) C = (C * (BigInt)(n - i)) / i;
tmp = C * (BigInt)(a[n-i]);
if(i&1)
negative = negative + tmp;
else
positive = positive + tmp;
}
BigInt ans = positive - negative;
ans.print();
}
return 0;
}