POJ3982 序列

问题链接：POJ3982 序列。

问题简述：参见上述链接。

问题分析：这只是一个大整数数列的问题，有一个大整数类就解决了。数列类似于斐波拉契数列，只是稍微复杂一些。

程序说明：这里使用自己的功能简洁的大整数类。这个类是参考其他人的程序改造的。

AC的C++语言程序如下：

/* POJ3982 序列 */#include <cstdio>#include <iostream>#include <string>#include <sstream>using namespace std;/* 无符号整数类，整数放在字符串中，可以用整数初始化。 * 只有以下功能： * 1.数组 * 2.加运算。 */class UBigInt {private:    string num;public:    UBigInt();    UBigInt(int n);    void setNumber(string s);    const string& getNumber(); // retrieves the number    UBigInt operator + (UBigInt b);private:    string add(string number1, string number2);    UBigInt& operator [] (int n);};UBigInt::UBigInt() { // empty constructor initializes zero    num = "0";}UBigInt::UBigInt(int n) {    stringstream ss;    string s;    ss << n;    ss >> s;    setNumber(s);}void UBigInt::setNumber(string s) {    num = s;}const string& UBigInt::getNumber() { // retrieves the number    return num;}UBigInt UBigInt::operator + (UBigInt b) {    UBigInt addition;    addition.setNumber( add(getNumber(), b.getNumber() ) );    return addition;}string UBigInt::add(string number1, string number2) {    string add = (number1.length() > number2.length()) ?  number1 : number2;    int diffLength = abs( (int) (number1.size() - number2.size()) );    if(number1.size() > number2.size())        number2.insert(0, diffLength, '0'); // put zeros from left    else// if(number1.size() < number2.size())        number1.insert(0, diffLength, '0');    char carry = 0;    for(int i=number1.size()-1; i>=0; --i) {        add[i] = (carry+(number1[i]-'0')+(number2[i]-'0')) + '0';        if(i != 0) {            if(add[i] > '9') {                add[i] -= 10;                carry = 1;            } else                carry = 0;        }    }    if(add[0] > '9') {        add[0]-= 10;        add.insert(0,1,'1');    }    return add;}UBigInt& UBigInt::operator [] (int n) {    return *(this + (n*sizeof(UBigInt)));}int main(){    UBigInt t[99+1];    int a, b, c;    while(scanf("%d%d%d", &a, &b, &c) != EOF) {        t[0] = a;        t[1] = b;        t[2] = c;        for(int i=3; i<=99; i++)            t[i] = t[i-3] + t[i-2] + t[i-1];        cout << t[99].getNumber() << endl;    }    return 0;}