HDU 4927 Series 1 (High Precision) 2014 multi-school training 6th

Series 1 Time Limit: 2000/1000 MS (Java/others) memory limit: 262144/262144 K (Java/Others)

Problem descriptionlet A be an integral series {A1, A2,..., }.

The zero-order series of A is a itself.

The first-order series of A is {b1, b2,..., Bn-1}, where bi = ai + 1-ai.

The ith-order series of A is the first-order series of its (I-1) Th-order series (2 <= I <= n-1 ).

Obviusly, The (n-1) Th-order series of A is a single integer. Given a, figure out that integer.
Inputthe 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.
The second line consists of N integers, describing A1, A2,..., An. (0 <= AI <= 105)
 
Outputfor each test case, output the required integer in a line.
Sample Input

231 2 341 5 7 2

 
Sample output

0-5

Question: give n numbers, execute n-1 merge, use the last number minus the previous number for each merge, and then ask what is the last number. Analysis: After analysis, it is not difficult to find that the final result is C (n-1, 0) * A [n]-C (n-1, 1) * A [n-1]… C (n-1, n-1) * A [1]. The sign signs are alternating between one positive and one negative. This operation is mainly for large numbers. During the competition, I used the string in C ++ to write a large number. After several hours of calling the number, I kept fighting until the end of the competition. While teammates quickly fell out of water using Java ,. After the competition, I found that only a few people wrote the question in C ++. It seems that it is really convenient to write large numbers in Java.

#include<cstdio>#include<cstring>using namespace std;#define MAXN 9999#define MAXSIZE 10#define DLEN 4class 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;}