A.Flipping Game 題目總結出來就是要求求出一個【i，j】區間，其中0的個數與1的個數是所有區間相差最大的。 題目有個trick，操作一定要執行,所以全是1的時候直接輸出n-1。 [cpp] int a[MAX]; int num0,num1; int main() { int n,i,j,k; int ans = 0; cin >> n; for(i=1; i<=n; i++) { scanf("%d",&a[i]); if(a[i] == 1) ans++; } if(ans == n) { cout << n-1 << endl; return 0; } int span = 0; for(i=1; i<=n; i++) { for(j=1; j<=n; j++) { num0 = 0; num1 = 0; for(k=i; k<=j; k++) { if(a[k] == 0) num0++; if(a[k] == 1) num1++; } span = max(span,num0 - num1); } } cout << span + ans << endl; return 0; } int a[MAX]; int num0,num1; int main() { int n,i,j,k; int ans = 0; cin >> n; for(i=1; i<=n; i++) { scanf("%d",&a[i]); if(a[i] == 1) ans++; } if(ans == n) { cout << n-1 << endl; return 0; } int span = 0; for(i=1; i<=n; i++) { for(j=1; j<=n; j++) { num0 = 0; num1 = 0; for(k=i; k<=j; k++) { if(a[k] == 0) num0++; if(a[k] == 1) num1++; } span = max(span,num0 - num1); } } cout << span + ans << endl; return 0; } B.Hungry Sequence special judge。。。所以打個素數表直接輸出即可 [cpp] int flag[MAX]; void prime() { flag[1] = 1; for(int i=2; i<=MAX; i++) { if(flag[i] == 0) { for(int j=2*i; j<=MAX; j+=i) flag[j] = 1; } } } int main() { int n,i; prime(); cin >> n; int cnt = 0; for(i=2;;i++) { if(flag[i] == 0) { cnt++; if(cnt == n) { cout << i << endl; break; } else { cout << i << ' '; } } } return 0; } int flag[MAX]; void prime() { flag[1] = 1; for(int i=2; i<=MAX; i++) { if(flag[i] == 0) { for(int j=2*i; j<=MAX; j+=i) flag[j] = 1; } } } int main() { int n,i; prime(); cin >> n; int cnt = 0; for(i=2;;i++) { if(flag[i] == 0) { cnt++; if(cnt == n) { cout << i << endl; break; } else { cout << i << ' '; } } } return 0; } C.Magic Five 快速冪，擴展歐幾裡得求逆元，熟練這兩個就好做了....可是我不熟 [cpp] char str[MAX]; __int64 k; int sum[MAX]; __int64 model(__int64 a,__int64 n,__int64 b) { __int64 t = a; __int64 ans = 1; while(n) { if(n & 1) { ans = ans * t % b; } n >>= 1; t = t*t % b; } return ans; } int main() { scanf("%s",str); scanf("%I64d",&k); int len = strlen(str); __int64 ans = 0; __int64 q = 0; for(int i=0; i<len; i++) { if(str[i] == '0' || str[i] == '5') { q = (q + model(2,i,mod)) % mod; } } ans = model(2,len*k,mod) % mod - 1; ans = (ans*model(model(2,len,mod) -1,mod - 2,mod)) % mod; ans = (q*ans) % mod; printf("%I64d\n",ans); return 0; } char str[MAX]; __int64 k; int sum[MAX]; __int64 model(__int64 a,__int64 n,__int64 b) { __int64 t = a; __int64 ans = 1; while(n) { if(n & 1) { ans = ans * t % b; } n >>= 1; t = t*t % b; } return ans; } int main() { scanf("%s",str); scanf("%I64d",&k); int len = strlen(str); __int64 ans = 0; __int64 q = 0; for(int i=0; i<len; i++) { if(str[i] == '0' || str[i] == '5') { q = (q + model(2,i,mod)) % mod; } } ans = model(2,len*k,mod) % mod - 1; ans = (ans*model(model(2,len,mod) -1,mod - 2,mod)) % mod; ans = (q*ans) % mod; printf("%I64d\n",ans); return 0; } D.Block Tower 帶有YY的爆搜，滿足讓一個聯通塊內只有一個blue房子其他都是red房子 [cpp] char map[MAX][MAX]; int vis[MAX][MAX]; struct node { int x,y; char dir; } p[1000005]; int dirx[4] = {1,-1,0,0}; int diry[4] = {0,0,1,-1}; int ans = 0 ,i,j; void dfs(int x,int y) { p[ans].dir = 'B'; p[ans].x = x; p[ans].y = y; ans++; vis[x][y] = 1; for(int k=0; k<4; k++) { int xx = x + dirx[k]; int yy = y + diry[k]; if(!vis[xx][yy] && map[xx][yy] == '.') dfs(xx,yy); } if(x != i || j != y){ p[ans].dir = 'D'; p[ans].x = x; p[ans].y = y; ans++; p[ans].dir = 'R'; p[ans].x =x; p[ans].y = y; ans++; } } int main() { int n,m; scanf("%d%d",&n,&m); for(i=0; i<n; i++) { scanf("%s",map[i]); } for(i=0; i<n; i++) { for(j=0; j<m; j++) { if(map[i][j] == '.' && !vis[i][j]) { dfs(i,j); } } } printf("%d\n",ans); for(i=0; i<ans; i++) { printf("%c %d %d\n",p[i].dir,p[i].x+1,p[i].y+1); } return 0; } char map[MAX][MAX]; int vis[MAX][MAX]; struct node { int x,y; char dir; } p[1000005]; int dirx[4] = {1,-1,0,0}; int diry[4] = {0,0,1,-1}; int ans = 0 ,i,j; void dfs(int x,int y) { p[ans].dir = 'B'; p[ans].x = x; p[ans].y = y; ans++; vis[x][y] = 1; for(int k=0; k<4; k++) { int xx = x + dirx[k]; int yy = y + diry[k]; if(!vis[xx][yy] && map[xx][yy] == '.') dfs(xx,yy); } if(x != i || j != y){ p[ans].dir = 'D'; p[ans].x = x; p[ans].y = y; ans++; p[ans].dir = 'R'; p[ans].x =x; p[ans].y = y; ans++; } } int main() { int n,m; scanf("%d%d",&n,&m); for(i=0; i<n; i++) { scanf("%s",map[i]); } for(i=0; i<n; i++) { for(j=0; j<m; j++) { if(map[i][j] == '.' && !vis[i][j]) { dfs(i,j); } } } printf("%d\n",ans); for(i=0; i<ans; i++) { printf("%c %d %d\n",p[i].dir,p[i].x+1,p[i].y+1); } return 0; }
