#include<iostream>
using namespace std;
int main()
{
int b[10][10],f=0,sum=0;
while(1){
for (int i=1;i<=9;i++)
for (int g=1;g<=9;g++)
scanf("%d",&b[i][g]);
for (int i=1;i<=9;i++){
for (int g=1;g<=9;g++)
sum+=b[i][g];
if (sum!=45)
{f=1;}else{sum=0;}}
if (f==1)
{cout<<"no\n";}else{cout<<"yes\n";}
}
}
TLE 請問有更有效率的解法嗎
#include
using namespace std;
int main()
{
int b[10][10],f=0,sum=0;
while(1){
for (int i=1;i<=9;i++)
for (int g=1;g<=9;g++)
scanf("%d",&b[i][g]);
for (int i=1;i<=9;i++){
for (int g=1;g<=9;g++)
sum+=b[i][g];
if (sum!=45)
{f=1;}else{sum=0;}}
if (f==1)
{cout<<"no\n";}else{cout<<"yes\n";}
}
}
TLE 請問有更有效率的解法嗎
善用EOF
while loop會是infinite loop