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},{47,...

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 uppersum...

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) { ...