我有每2阵列5整数,我怎么能比较两种阵列,使阵列A比阵列B大,下面是我的code在那里我得到的回报0:
的#include<的iostream>
使用名字空间std;
诠释的main()
{
const int的AMAX = 5,BMAX = 6;
INT I;
布尔C1 = TRUE,C2 = FALSE;
INT A〔AMAX] = {1,2,3,4,5};
INT B〔BMAX] = {6,7,8,9,1};
对于(i = 0; I< BMAX;我++)
如果(B [我] == A [1])
COUT<< C1<< ENDL;
其他
COUT<< C2<< ENDL;
返回0;
}
我在想什么吗?
更新:
的#include<的iostream>
使用名字空间std;
诠释的main(){
诠释一个[] = {6,7,29};
INT B〔] = {3,2,11};
INT ACC1 = 0;
INT ACC2 = 0;
的for(int i = 0;我3;;我++){
ACC1 + = A [1];
}
为(诠释J = 0; J&所述; 3; J ++){
ACC2 + = B [J]。
}
如果(ACC1< ACC2){
的printf(数组b大于数组b);
}
其他{
的printf(数组b大于阵列A);
}
返回0;
}
解决方案
如果你想比较哪些的金额的数组中的元素(即确定是否所有元素的总和 A
大于所有元素的 B
)的总和,因为这个问题的标题似乎在暗示,这应该做些什么你想要的:
的#include<的iostream>
#包括<算法>
#包括<数字>
#包括<了iomanip>
诠释的main()
{
使用名字空间std;
INT A [] = {2,4,1,5,9};
INT B〔] = {9,12,32,43,23};
INT SUMA =累加(开始(A),端(A),0);
INT sumB =累加(开始(B),终端(B),0);
COUT<< boolalpha<< (SUMA> sumB);
返回0;
}
如果你不使用C ++ 11,这里是写在同等功能的程序C ++ 03:
的#include<的iostream>
#包括<数字>
诠释的main()
{
使用名字空间std;
INT A [] = {2,4,1,5,9};
INT B〔] = {9,12,32,43,23};
INT SUMA =累加(A,A +的sizeof(A)/的sizeof(INT),0);
INT sumB =累加(B,B +的sizeof(B)/的sizeof(INT),0);
COUT<< ((SUMA> sumB)真:假);
返回0;
}
i have 2 arrays each with 5 integers, how can i compare both of array so that array A is larger than array B, following are my code where i get return 0:
#include <iostream>
using namespace std;
int main()
{
const int AMAX = 5, BMAX = 6;
int i;
bool c1 = true, c2 = false;
int A[AMAX] = { 1, 2, 3, 4, 5 };
int B[BMAX] = { 6, 7,8, 9, 1};
for (i = 0; i < BMAX; i++)
if (B[i] == A[i])
cout << c1 << endl;
else
cout << c2 << endl;
return 0;
}
What am i missing here?
update:
#include <iostream>
using namespace std;
int main(){
int a[] = {6,7,29};
int b[] = {3,2,11};
int acc1=0;
int acc2 = 0;
for (int i=0;i<3;i++){
acc1+=a[i];
}
for(int j=0;j<3;j++){
acc2+=b[j];
}
if(acc1 < acc2){
printf("Array B is greater than Array B ");
}
else{
printf("Array B greater than Array A");
}
return 0;
}
解决方案
If what you want to compare are the sums of the elements in the array (i.e. determine whether the sum of all elements in A
is greater than the sum of all elements in B
), as the question's title seems to imply, this should do what you want:
#include <iostream>
#include <algorithm>
#include <numeric>
#include <iomanip>
int main()
{
using namespace std;
int A[] = { 2, 4, 1, 5, 9 };
int B[] = { 9, 12, 32, 43, 23};
int sumA = accumulate(begin(A), end(A), 0);
int sumB = accumulate(begin(B), end(B), 0);
cout << boolalpha << (sumA > sumB);
return 0;
}
In case you are not using C++11, here is the equivalent program written in C++03:
#include <iostream>
#include <numeric>
int main()
{
using namespace std;
int A[] = { 2, 4, 1, 5, 9 };
int B[] = { 9, 12, 32, 43, 23};
int sumA = accumulate(A, A + sizeof(A) / sizeof(int), 0);
int sumB = accumulate(B, B + sizeof(B) / sizeof(int), 0);
cout << ((sumA > sumB) ? "true" : "false");
return 0;
}
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