问题可以在这里找到：http://projecteuler.net/problem=11
昨天我看到了Euler项目，我很喜欢它。我制作了一个代码来解决问题11，但由于某种原因，我无法看到答案是错误的。
有人可以阅读我的代码并提出建议吗？
#include <iostream>
using namespace std;
int input [20][20] = {
{8,2,22,97,38,15,0,40,0,75,4,5,7,78,52,12,50,77,91,8},
{49,49,99,40,17,81,18,57,60,87,17,40,98,43,69,48,4,56,62,0},
{81,49,31,73,55,79,14,29,93,71,40,67,53,88,30,3,49,13,36,65},
{52,70,95,23,4,60,11,42,69,24,68,56,1,32,56,71,37,2,36,91},
{22,31,16,71,51,67,63,89,41,92,36,54,22,40,40,28,66,33,13,80},
{24,47,32,60,99,3,45,2,44,75,33,53,78,36,84,20,35,17,12,50},
{32,98,81,28,64,23,67,10,26,38,40,67,59,54,70,66,18,38,64,70},
{67,26,20,68,2,62,12,20,95,63,94,39,63,8,40,91,66,49,94,21},
{24,55,58,5,66,73,99,26,97,17,78,78,96,83,14,88,34,89,63,72},
{21,36,23,9,75,0,76,44,20,45,35,14,0,61,33,97,34,31,33,95},
{78,17,53,28,22,75,31,67,15,94,3,80,4,62,16,14,9,53,56,92},
{16,39,5,42,96,35,31,47,55,58,88,24,0,17,54,24,36,29,85,57},
{86,56,0,48,35,71,89,7,5,44,44,37,44,60,21,58,51,54,17,58},
{19,80,81,68,5,94,47,69,28,73,92,13,86,52,17,77,4,89,55,40},
{4,52,8,83,97,35,99,16,7,97,57,32,16,26,26,79,33,27,98,66},
{88,36,68,87,57,62,20,72,3,46,33,67,46,55,12,32,63,93,53,69},
{4,42,16,73,38,25,39,11,24,94,72,18,8,46,29,32,40,62,76,36},
{20,69,36,41,72,30,23,88,34,62,99,69,82,67,59,85,74,4,36,16},
{20,73,35,29,78,31,90,1,74,31,49,71,48,86,81,16,23,57,5,54},
{1,70,54,71,83,51,54,69,16,92,33,48,61,43,52,1,89,19,67,48}
};
/* if element is a[i][j], its diagonal to the right is a[i+1][j+1]
if element is a[i][j], its diagonal to the left is a[i-1][j-1]
IF element is a[i][j], the element below it is a[i+1][j]
IF element is a[i][j], the element on right of it is a[i][j+1]
*/
/*
approach:
=> find the greatest product of diagonals
=> find the greatest prodct of elements below each other
=> find the greatest prodct of element on right of each other
=> compare them
*/
int main(void)
{
unsigned long hr=0,hb=0,hdr=0, hdl=0;
//hr is the highest product on the right side traversal
//similraly for others
//for diagonal right
for(int i =0; i<17;i++) //going from 0 to 16 so that we dont land up beyond the array for (i+1),(j+1) etc
{
for(int j=0;j<17;j++)
{
if(input[i][j]*input[i+1][j+1]*input[i+2][j+2]*input[i+3][j+3] > hdr)
{
hdr = input[i][j]*input[i+1][j+1]*input[i+2][j+2]*input[i+3][j+3];
}
}
}
for(int i =19; i>=3;i--) //for diagonal left
{
for(int j=19;j>=3;j--)
{
if(input[i][j]*input[i-1][j-1]*input[i-2][j-2]*input[i-3][j-3] > hdl)
{
hdl = input[i][j]*input[i-1][j-1]*input[i-2][j-2]*input[i-3][j-3];
}
}
}
for(int i =0; i<17;i++) //for elements below each other
{
for(int j=0;j<20;j++)
{
if(input[i][j]*input[i+1][j]*input[i+2][j]*input[i+3][j] > hb)
{
hb = input[i][j]*input[i+1][j]*input[i+2][j]*input[i+3][j];
}
}
}
for(int i =0; i<20;i++) //on right
{
for(int j=0;j<17;j++)
{
if(input[i][j]*input[i][j+1]*input[i][j+2]*input[i][j+3] > hr)
{
hr = input[i][j]*input[i][j+1]*input[i][j]*input[i][j+3];
}
}
}
if(hdr>hb && hdr > hr && hdr>hdl )
{
cout<<hdr<<endl;
}
else if (hb > hdr && hb > hr && hb>hdl)
{
cout<<hb<<endl;
}
else if(hr>hb && hr> hdr && hr > hdl)
{
cout<<hr<<endl;
}
else
{
cout<<hdl<<endl;
}
return 0;
}
答案 0 :(得分：0)
这是正确的代码。我建议任何人尝试这个问题都需要一些时间来实际用铅笔和纸来看模式。这个问题本身很容易。
#include <iostream>
using namespace std;
int input [20][20] ={{8,2,22,97,38,15,0,40,0,75,4,5,7,78,52,12,50,77,91,8}, {49,49,99,40,17,81,18,57,60,87,17,40,98,43,69,48,4,56,62,0},
{81,49,31,73,55,79,14,29,93,71,40,67,53,88,30,3,49,13,36,65},
{52,70,95,23,4,60,11,42,69,24,68,56,1,32,56,71,37,2,36,91},
{22,31,16,71,51,67,63,89,41,92,36,54,22,40,40,28,66,33,13,80},
{24,47,32,60,99,3,45,2,44,75,33,53,78,36,84,20,35,17,12,50},
{32,98,81,28,64,23,67,10,26,38,40,67,59,54,70,66,18,38,64,70},
{67,26,20,68,2,62,12,20,95,63,94,39,63,8,40,91,66,49,94,21},
{24,55,58,5,66,73,99,26,97,17,78,78,96,83,14,88,34,89,63,72},
{21,36,23,9,75,0,76,44,20,45,35,14,0,61,33,97,34,31,33,95},
{78,17,53,28,22,75,31,67,15,94,3,80,4,62,16,14,9,53,56,92},
{16,39,5,42,96,35,31,47,55,58,88,24,0,17,54,24,36,29,85,57},
{86,56,0,48,35,71,89,7,5,44,44,37,44,60,21,58,51,54,17,58},
{19,80,81,68,5,94,47,69,28,73,92,13,86,52,17,77,4,89,55,40},
{4,52,8,83,97,35,99,16,7,97,57,32,16,26,26,79,33,27,98,66},
{88,36,68,87,57,62,20,72,3,46,33,67,46,55,12,32,63,93,53,69},
{4,42,16,73,38,25,39,11,24,94,72,18,8,46,29,32,40,62,76,36},
{20,69,36,41,72,30,23,88,34,62,99,69,82,67,59,85,74,4,36,16},
{20,73,35,29,78,31,90,1,74,31,49,71,48,86,81,16,23,57,5,54},
{1,70,54,71,83,51,54,69,16,92,33,48,61,43,52,1,89,19,67,48}};
/* if element is a[i][j], its diagonal to the right is a[i+1][j+1]
if element is a[i][j], its diagonal to the left is a[i+1][j-1]
IF element is a[i][j], the element below it is a[i+1][j]
IF element is a[i][j], the element on right of it is a[i][j+1]
*/
/*
approach:
=> find the greatest product of diagonals
=> find the greatest prodct of elements below each other
=> find the greatest prodct of element on right of each other
=> compare them
*/
int main(void)
{
unsigned long hr=0,hb=0,hdr=0, hdl=0;
//hr is the highest product on the right side traversal
//similraly for others
//for diagonal right
for(int i =0; i<17;i++) //going from 0 to 16 so that we dont land up beyond the array for (i+1),(j+1) etc
{
for(int j=0;j<17;j++)
{
if(input[i][j]*input[i+1][j+1]*input[i+2][j+2]*input[i+3][j+3] > hdr)
{
hdr = input[i][j]*input[i+1][j+1]*input[i+2][j+2]*input[i+3][j+3];
}
}
}
for(int i =0; i<17;i++) //for diagonal left
{
for(int j=19;j>=3;j--)
{
if(input[i][j]*input[i+1][j-1]*input[i+2][j-2]*input[i+3][j-3] > hdl)
{
hdl = input[i][j]*input[i+1][j-1]*input[i+2][j-2]*input[i+3][j-3];
}
}
}
for(int i =0; i<17;i++) //for elements below each other
{
for(int j=0;j<20;j++)
{
if(input[i][j]*input[i+1][j]*input[i+2][j]*input[i+3][j] > hb)
{
hb = input[i][j]*input[i+1][j]*input[i+2][j]*input[i+3][j];
}
}
}
for(int i =0; i<20;i++) //on right
{
for(int j=0;j<17;j++)
{
if(input[i][j]*input[i][j+1]*input[i][j+2]*input[i][j+3] > hr)
{
hr = input[i][j]*input[i][j+1]*input[i][j]*input[i][j+3];
}
}
}
if(hdr>hb && hdr > hr && hdr>hdl )
{
cout<<hdr<<endl;
}
else if (hb > hdr && hb > hr && hb>hdl)
{
cout<<hb<<endl;
}
else if(hr>hb && hr> hdr && hr > hdl)
{
cout<<hr<<endl;
}
else
{
cout<<hdl<<endl;
}
return 0;
}