Accumulate函式學習、

當以迭代器first和last及值init作為引數呼叫時

accumulate(first,last, init);將把init 和從 first 到last 指向的值進行累加，並返回累加得到的和，但不包括last指向的值。通過例項學習accumulate 函式。

#include<iostream>
#include<vector>
#include<numeric>
using namespace std;
int main()
{
   cout<<"Demonstratint the accumulatefunction."<<endl;
   int x[5] = {2,3,5,7,11};
    vector<int>vectorl(&x[0],&x[5]);
   int sum = accumulate(vectorl.begin(),vectorl.end(),0);
cout<<"sum ="<<sum<<endl;
//accumulate 直接對陣列進行操作
   int sum1 = accumulate(&x[0],&x[5],0);
   cout<<"sum1 = "<<sum1<<endl;
double y[5] = {2.1,3.123,5.3,7.0,11.0};
// accumulate 對浮點型資料相加。
   vector<double>vectorl1(&y[0],&y[5]);
   double sum2 = accumulate(vectorl1.begin(),vectorl1.end(),0.0);
   cout<<"sum2 = "<<sum2<<endl;
   return 0;
}
 

結果是

Demonstratint the accumulate function.

sum = 28

sum1 = 28

sum2 = 28.523

accumulate 計算連乘，

#include<iostream>
#include<vector>
#include<cassert>
#include<numeric>
using namespace std;
int mult(int x, int y)
{
   return x * y;
}
int main()
{
   int x[5] = {2, 3, 5, 7, 11};
   vector<int>vectorl(&x[0], &x[5]);
   int product = accumulate(vectorl.begin(), vectorl.end(), 1, mult);
   cout<<"product = "<<product<<endl;
   return 0;
}
 

結果為

product = 2310

這裡傳遞給accumulate 函式一個普通函式 mult ，實際上傳遞的是該函式的地址。

用multiplies 和 accumulate 求連乘

#include<iostream>
#include<vector>
#include<numeric>
#include<functional>
using namespace std;
int main()
{
   int x[5] = {2, 3, 5, 7, 11};
   vector<int>vectorl(&x[0], &x[5]);
   int product = accumulate(vectorl.begin(), vectorl.end(), 1,multiplies<int>());
   cout<<"product = "<<product<<endl;
   return 0;
}
 

結果為

product = 2310

通過multiplies<int>() 呼叫了以int型別例項化的multiplies類的預設建構函式。

multiplies標頭檔案為#include <functional>

一個例項在 http://www.linuxidc.com/Linux/2015-04/116423.htm 搬過來學習。

#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中個元素的值為:\n ( " ;
   for ( Iter1 = v1.begin( ) ; Iter1 != v1.end( ) ; Iter1++ )
      cout << *Iter1 << " ";
   cout << ")." << endl;


   // accumulate函式的第一個功能，求和
   int total;
   total = accumulate ( v1.begin ( ) , v1.end ( ) , 0 );


   cout << "整數從1到20的和為: "
        << total << "." << endl;


   // 構造一個前n項和的向量
   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項和分別為:\n ( " ;
   for ( Iter2 = v2.begin( ) ; Iter2 != v2.end( ) ; Iter2++ )
      cout << *Iter2 << " ";
   cout << ")." << endl << endl;


   // accumulate函式的第二個功能，計算連乘積
   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的初始值分別為:\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 << "整數1到10的連乘積為: "
        << ptotal << "." << endl;


   // 構造一個前n項積的向量
   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項積分別為:\n ( " ;
   for ( Iter4 = v4.begin( ) ; Iter4 != v4.end( ) ; Iter4++ )
      cout << *Iter4 << " ";
   cout << ")." << endl;
}

剛開始學習STL，當做記筆記了！！！