OJ Board

Alexander Kozlov wrote:Try this input:

1
4
8
0 1
1 2
2 3
3 0
2 4
4 2
3 4
4 3

This is not cactus, but for every node in-degree equals to out-degree.
Am I right?

is that we can slove in biconnective algorithm?
biconnective can be to find the SCC and back/forward/tree edges

You can compare me code
My code can do the right way

[c]
#include<stdio.h>

int main(void)
{
int input=0;

while(1==scanf("%d", &input))
{
if(0==input)
return 0;
else
{
int array[25][2]=
{{2,0},{3,0},{5,0},{7,0},{11,0}
,{13,0},{17,0},{19,0},{23,0},{29,0}
,{31,0},{37,0},{41,0},{43,0 ...

The WA problem confused me at all!!! :evil:
The sample input which my program can do in right way, but the judge gave me WA......

there is the code
[c]
#include<stdio.h>

void WriteSumAndPath(long array[], long *upper, long *middle, long *down)
{
long smallest=*middle, path=*(middle+1);
long ...

I got it!!!

sorry

it is WA

[c]
#include<stdio.h>

int main(void)
{
/* R = B^P mod M */
unsigned long B=0, P=0;
unsigned int M=0;

while(3==scanf("%d %d %d", &B, &P, &M))
{
int mod=1, i=0;

if(0==B%M) //best case
{
printf("0\n");
continue;
}
else if(B>M) //let B < M
B%=M;

while(1<P)
{
if(1==P%2)
{
mod*=B ...