C++ STL算法之accumulate函数

1. 介绍

　　用来计算特定范围内（包括连续的部分和初始值）所有元素的和，除此之外，还可以用指定的二进制操作来计算特定范围内的元素结果。其头文件在numeric中。

　　accumulate原函数声明定义如下：

template<class InputIterator, class Type>
   Type accumulate(
      InputIterator _First,
      InputIterator _Last,
      Type _Val
   );
template<class InputIterator, class Type, class Fn2>
   Type accumulate(
      InputIterator _First,
      InputIterator _Last,
      Type _Val,
      BinaryOperation _Binary_op
   );

参数说明。

但看说明也许不太容易懂，我们通过例子更具体的解释下这个函数。

2. 应用举例

#include <vector>
#include <numeric>
#include <functional>
#include <iostream>

using namespace std;

int main( )
{

   vector <int> v1, v2( 20 );
   vector <int>::iterator Iter1, Iter2;

   int i;
   for ( i = 1 ; i < 21 ; i++ )
   {
      v1.push_back( i );
   }

   cout << "最初向量v1中个元素的值为:\n ( " ;
   for ( Iter1 = v1.begin( ) ; Iter1 != v1.end( ) ; Iter1++ )
      cout << *Iter1 << " ";
   cout << ")." << endl;

   // accumulate函数的第一个功能，求和
   int total;
   total = accumulate ( v1.begin ( ) , v1.end ( ) , 0 );

   cout << "整数从1到20的和为: "
        << total << "." << endl;

   // 构造一个前n项和的向量
   int j = 0, partotal;
   for ( Iter1 = v1.begin( ) + 1; Iter1 != v1.end( ) + 1 ; Iter1++ )
   {
      partotal = accumulate ( v1.begin ( ) , Iter1 , 0 );
      v2 [ j ] = partotal;
      j++;
   }

   cout << "前n项和分别为:\n ( " ;
   for ( Iter2 = v2.begin( ) ; Iter2 != v2.end( ) ; Iter2++ )
      cout << *Iter2 << " ";
   cout << ")." << endl << endl;

   // accumulate函数的第二个功能，计算连乘积
   vector <int> v3, v4( 10 );
   vector <int>::iterator Iter3, Iter4;

   int s;
   for ( s = 1 ; s < 11 ; s++ )
   {
      v3.push_back( s );
   }

   cout << "向量v3的初始值分别为:\n ( " ;
   for ( Iter3 = v3.begin( ) ; Iter3 != v3.end( ) ; Iter3++ )
      cout << *Iter3 << " ";
   cout << ")." << endl;

   int ptotal;
   ptotal = accumulate ( v3.begin ( ) , v3.end ( ) , 1 , multiplies<int>( ) );

   cout << "整数1到10的连乘积为: "
        << ptotal << "." << endl;

   // 构造一个前n项积的向量
   int k = 0, ppartotal;
   for ( Iter3 = v3.begin( ) + 1; Iter3 != v3.end( ) + 1 ; Iter3++ ) {
      ppartotal = accumulate ( v3.begin ( ) , Iter3 , 1 , multiplies<int>( ) );
      v4 [ k ] = ppartotal;
      k++;
   }

   cout << "前n项积分别为:\n ( " ;
   for ( Iter4 = v4.begin( ) ; Iter4 != v4.end( ) ; Iter4++ )
      cout << *Iter4 << " ";
   cout << ")." << endl;
}

编译运行，看一下输出结果：