//CodeChef RRMATRIX
//分析：(x-1)*m+y == (y-1)*n+x ==> (x-1)*(m-1) == (y-1)*(n-1) ==> (x-1) == (y-1)*(n-1)/(m-1);(n-1)/(m-1)约分后分母还剩余一个(m-1)/gcd(n-1,m-1),(y-1)是(m-1)/gcd(n-1,m-1)倍数的情况共有(m-1)/((m-1)/gcd(n-1,m-1))+1种，即gcd(n-1,m-1)+1
1 #include"iostream" 2 #include"cstdio" 3 using namespace std; 4 int main() 5 { 6 int T,n,m,temp; 7 scanf("%d",&T); 8 while(T--) { 9 scanf("%d%d",&n,&m); 10 if(n==1) { //n==1 || m==1为两种特殊情况，需特判 11 printf("%d ",m); 12 continue; 13 } 14 if(m==1) { 15 printf("%d ",n); 16 continue; 17 } 18 --n; 19 --m; 20 while(m) { 21 temp = n%m; 22 n = m; 23 m = temp; 24 } 25 printf("%d ",n+1); 26 } 27 return 0; 28 }
//CodeForces 24A
//分析：对整个“有向环”做一趟dfs染色即可
//因为是入栈后标记，所以访问过的点一定会被标记，由于需要回到起点，所以记录下最后一次访问过的点last_point；相对应的，入栈前标记的话，首次访问的点不会被标记，用这种方法写本题的话，由于 1 点未被标记，所以可能会在一个不希望的时机去第二次访问 1 点，就需要删除一条从 1 点出发的出边，和另一条回到 1 点的回边，而且对于只有两个点的图还需要特判，比较麻烦
1 #include"iostream" 2 #include"cstdio" 3 #include"cstring" 4 using namespace std; 5 const int maxn = 110; 6 int tot,first[maxn<<1],next[maxn<<1],destination[maxn<<1],weight[maxn<<1]; 7 int n,res,last_point; 8 bool vis[maxn]; 9 void dfs(int point) 10 { 11 vis[point] = 1; 12 last_point = point; 13 int edge = first[point]; 14 while(edge) { 15 if(!vis[destination[edge]]) { 16 res += weight[edge]; 17 dfs(destination[edge]); 18 } 19 edge = next[edge]; 20 } 21 } 22 23 24 int main() 25 { 26 int i,u,v,w,sum = 0; 27 res = tot = 0; 28 scanf("%d",&n); 29 for(i = 1; i<=n; i++) { 30 scanf("%d%d%d",&u,&v,&w); 31 next[++tot] = first[u]; 32 first[u] = tot; 33 destination[tot] = v; 34 weight[tot] = 0; 35 next[++tot] = first[v]; 36 first[v] = tot; 37 destination[tot] = u; 38 weight[tot] = w; 39 sum += w; 40 } 41 dfs(1); 42 int edge = first[last_point]; 43 while(destination[edge]!=1) { 44 edge = next[edge]; 45 } 46 res += weight[edge]; 47 printf("%d ",res<sum-res?res:sum-res); 48 return 0; 49 }
1 #include"iostream" 2 #include"cstdio" 3 #include"cstring" 4 using namespace std; 5 const int maxn = 110; 6 int tot,first[maxn<<1],next[maxn<<1],destination[maxn<<1],weight[maxn<<1]; 7 int n,res,last_point; 8 bool vis[maxn]; 9 int dfs(int point) 10 { 11 vis[point] = 1; 12 last_point = point; 13 int edge = first[point]; 14 while(edge) { 15 if(!vis[destination[edge]]) { 16 return weight[edge]+dfs(destination[edge]); 17 } 18 edge = next[edge]; 19 } 20 return 0; //注意返回0 21 } 22 23 24 int main() 25 { 26 int i,u,v,w,sum = 0; 27 tot = 0; 28 scanf("%d",&n); 29 for(i = 1; i<=n; i++) { 30 scanf("%d%d%d",&u,&v,&w); 31 next[++tot] = first[u]; 32 first[u] = tot; 33 destination[tot] = v; 34 weight[tot] = 0; 35 next[++tot] = first[v]; 36 first[v] = tot; 37 destination[tot] = u; 38 weight[tot] = w; 39 sum += w; 40 } 41 res = dfs(1); 42 int edge = first[last_point]; 43 while(destination[edge]!=1) { 44 edge = next[edge]; 45 } 46 res += weight[edge]; 47 printf("%d ",res<sum-res?res:sum-res); 48 return 0; 49 }