Algorithm常用函数（1）

1.求和——accumulate

 1 #include<iostream>
 2 #include<cstring>
 3 #include<cstdio>
 4 
 5 #include<numeric>//accumulate
 6 using namespace std;
 7 
 8 int main()
 9 {
10     int all[100] = {2,2,3,4,5,6,7,8,9,10};
11     int sum = accumulate(all, all + 3, 1);//all的前三个数相加，和的初始值是1
12     printf("%d", sum);
13 
14     return 0;
15 }

2.二分查找—— binary_search

 1 #include<algorithm>
 2 #include<numeric>
 3 
 4 using namespace std;
 5 
 6 int main()
 7 {
 8     int all[100] = {0, 1, 2, 3, 4, 5, 89, 104, 300, 600, 9000};
 9     int a;
10 
11     while(scanf_s("%d", &a, 1) != EOF)
12     {
13         if(binary_search(all, all + 10, a))//在all的前十个元素中搜索a
14             printf_s("Have
");
15         else
16             printf_s("Not have
");
17     }
18 
19     return 0;
20 }

3.拷贝函数——copy与copy_backward

 1 #include<cstdio>
 2 #include<iostream>
 3 #include<cstring>
 4 
 5 ///copy
 6 #include<algorithm>
 7 #include<numeric>
 8 using namespace std;
 9 
10 class N
11 {
12 public:
13     int a;
14     int b;
15     int c;
16 }all1[100], all2[100];
17 
18 int main()
19 {
20     all1[0].a = 1;
21     all1[0].b = 2;
22     all1[0].c = 3;
23 
24     all1[1].a = 4;
25     all1[1].b = 5;
26     all1[1].c = 6;
27 
28     copy(all1, all1 + 2, all2);//copy_backward
29 
30     for(int i = 0 ; i < 2 ; i ++)
31     {
32         printf_s("%d	%d	%d
", all2[i].a, all2[i].b, all2[i].c);
33     }
34 
35     return 0;
36 }

4. 计数——count

 1 #include<cstdio>
 2 #include<cstring>
 3 #include<iostream>
 4 
 5 //count
 6 #include<algorithm>
 7 #include<numeric>
 8 
 9 using namespace std;
10 
11 int main()
12 {
13     int all[11] = {0, 1, 0, 0, 1, 0, 1, 5, 1, 0, 0};
14 
15     int num = count(all, all + 8, 5);//数一下all的前八个数里面1的个数
16     printf_s("%d
", num);
17 
18     return 0;
19 }

5. 条件计数——count_if

 1 #include<cstdio>
 2 #include<iostream>
 3 #include<cstring>
 4 
 5 //count_if
 6 #include<algorithm>
 7 #include<numeric>
 8 
 9 using namespace std;
10 
11 int main()
12 {
13     int all[11] = {0, 1, 2, 3, 4, 5, 6, 4, 5, 6, 10};
14     int sum = count_if(all, all + 11,bind2nd(equal_to<int>(), 5));//这种方法貌似是被遗弃鸟~~
15 
16     printf_s("%d", sum);
17 
18     return 0;
19 }

6. 相等——equal

 1 #include<cstdio>
 2 #include<iostream>
 3 #include<cstring>
 4 
 5     //equal
 6 #include<algorithm>
 7 #include<numeric>
 8 
 9     using namespace std;
10 
11 int main()
12 {
13     int all[11]={0,1,2,3,4,5,6,7,8,9,10};
14     int all2[11]={0,1,2,3,4,5,6,7,8,9,11};
15     if(equal(all, all + 10, all2))//比较all与all2的前10个元素是否相等，可以扩展到对象是否相等
16         printf_s("equal
");
17     else
18         printf_s("not equal
");
19 
20     return 0;
21 }

7. 填充——fill，fill_n

#include<cstdio>
#include<iostream>
#include<cstring>

//fill，fill_n
#include<algorithm>
#include<numeric>

using namespace std;

int main()
{
    int all[11] = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10};

    //fill(all, all + 11 , 100);//将all的前11个元素填充成100
    fill_n(all, 5, 100);//前5个元素填充成100，其他不变
    printf_s("%d
", all[4]);
    printf_s("%d
", all[5]);
}

8. 查找——find，find_end，find_first_of，find_if

 1 #include<cstdio>
 2 #include<cstring>
 3 #include<iostream>
 4 
 5 //find，find_end，find_first_of，find_if
 6 #include<algorithm>
 7 #include<numeric>
 8 
 9 using namespace std;
10 
11 int main()
12 {
13     int all[14]={0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 4, 5, 6};
14     int *a = find(all, all + 11, 8);//在all的前11个元素中查找8，如果存在，则返回8所对应的指针
15     printf_s("%d
", *a);
16 
17     int all2[4] = {4, 5, 6};
18     a = find_end(all, all + 14, all2 + 0 , all2 + 3);//找到字串all2在all中最后出现的初始位置
19     printf_s("位置：%d
", a - all);
20 
21     a = find_first_of(all, all + 14, all2 + 0 , all2 + 3);//找到字串all2在all中第一次出现的初始位置
22     printf_s("位置：%d
", a - all);
23 
24 
25     int *sum=find_if(all,all+10,bind2nd(greater_equal<int> (), 5));//这种方法被遗弃了
26     printf("%d",sum-all);
27 
28     return 0;
29 }