标签：取数游戏 技术分享 oid noip play getchar log scan namespace
怎么说呢
#include<iostream> #include<cstdio> #include<algorithm> using namespace std; const int N = 5050; int n, l, pos[N], maxx = 0, minn = 0; int main(){ scanf("%d%d", &l, &n); for(int i = 1; i <= n; i++) scanf("%d", &pos[i]); sort(pos + 1, pos + n + 1); for(int i = 1; i <= n; i++) if(pos[i] <= (1 + l) / 2) maxx = max(maxx, l - pos[i] + 1), minn = max(pos[i], minn); else maxx = max(maxx, pos[i]), minn = max(minn, l - pos[i] + 1); printf("%d %d", minn, maxx); return 0; }
多维dp的应用
code传纸条
#include<iostream> #include<cstdio> #include<cstdlib> #include<cstring> #include<algorithm> #include<cmath> using namespace std; const int N = 60; int n, m, a[N][N], dp[N][N][N][N]; inline int myMax(int u, int v, int x, int y){ int ret = u; ret = max(ret, v); ret = max(ret, x); ret = max(ret, y); return ret; } int main(){ scanf("%d%d", &m, &n); for(int i = 1; i <= m; i++) for(int j = 1; j <= n; j++) scanf("%d", &a[i][j]); dp[1][1][1][1] = a[1][1]; for(int i = 1; i <= m; i++) for(int j = 1; j <= n; j++) for(int k = 1; k <= m; k++) for(int l = 1; l <= n; l++){ if(i == 1 && j == 1) continue; dp[i][j][k][l] = myMax(dp[i - 1][j][k - 1][l], dp[i][j - 1][k - 1][l], dp[i - 1][j][k][l - 1], dp[i][j - 1][k][l - 1]) + a[i][j] + a[k][l]; if(i == k && j == l) dp[i][j][k][l] -= a[i][j]; } printf("%d", dp[m][n][m][n]); return 0; }
code方格取数
#include<iostream> #include<cstdio> #include<cstdlib> #include<cstring> #include<algorithm> #include<cmath> using namespace std; const int N = 15; int n, a[N][N], dp[N][N][N][N]; inline int myMax(int u, int v, int x, int y){ int ret = u; ret = max(ret, v); ret = max(ret, x); ret = max(ret, y); return ret; } int main(){ scanf("%d", &n); int x, y, w; while(scanf("%d%d%d", &x, &y, &w), x + y + w) a[x][y] = w; // dp[1][1][1][1] = a[1][1]; for(int i = 1; i <= n; i++) for(int j = 1; j <= n; j++) for(int k = 1; k <= n; k++) for(int l = 1; l <= n; l++){ // if(i == k && j == l && i != n && j != n) continue; dp[i][j][k][l] = myMax(dp[i - 1][j][k - 1][l], dp[i][j - 1][k - 1][l], dp[i - 1][j][k][l - 1], dp[i][j - 1][k][l - 1]) + a[i][j] + a[k][l]; if(i == k && j == l ) dp[i][j][k][l] -= a[i][j]; } printf("%d", dp[n][n][n][n]); return 0; }
问题识破 + dp + 高精度
#include<iostream> #include<cstdio> #include<string> #include<cstring> using namespace std; inline int read(){ int i = 0, f = 1; char ch = getchar(); for(; (ch < ‘0‘ || ch > ‘9‘) && ch != ‘-‘; ch = getchar()); if(ch == ‘-‘) f = -1, ch = getchar(); for(; ch >= ‘0‘ && ch <= ‘9‘; ch = getchar()) i = (i << 3) + (i << 1) + (ch - ‘0‘); return i * f; } const int N = 100; int n, m; struct bign{ int len, s[N]; bign():len(1){memset(s, 0, sizeof s);} inline void clear(){ while(len > 1 && s[len] == 0) len--; } inline bign operator * (const bign &u) const{ bign ret; ret.len = len + u.len + 10; for(int i = 1; i <= len; i++) for(int j = 1; j <= u.len; j++) ret.s[i + j - 1] += s[i] * u.s[j]; for(int i = 1; i <= ret.len; i++) if(ret.s[i] >= 10){ ret.s[i + 1] += ret.s[i] / 10; ret.s[i] %= 10; } ret.clear(); return ret; } inline void print(){ clear(); for(int i = len; i >= 1; i--) putchar(s[i] + ‘0‘); } inline bool operator > (const bign &u) const{ if(len != u.len) return len > u.len; for(int i = len; i >= 1; i--) if(s[i] != u.s[i]) return s[i] > u.s[i]; return false; } inline bign operator + (const bign &u) const{ bign ret; ret.len = 0; int i, g; for(i = 1, g = 0; g || i <= max(len, u.len); i++){ ret.s[++ret.len] = g; if(i <= len) ret.s[ret.len] += s[i]; if(i <= u.len) ret.s[ret.len] += u.s[i]; g = ret.s[ret.len] / 10; ret.s[ret.len] %= 10; } ret.clear(); return ret; } }pow2[N][N], f[N][N], big2, ans, ret, a[N], big0; inline bign newBign(int x){ bign ret; ret.len = 0; while(x){ ret.s[++ret.len] = x % 10; x /= 10; } if(ret.len == 0) ret.len = 1; return ret; } inline void initPow(){ for(int i = 1; i <= m; i++) for(int j = 1; j <= m; j++) pow2[i][j] = pow2[i][j - 1] + pow2[i][j - 1]; } inline bign max(bign a, bign b){ return a > b ? a : b; } int main(){ n = read(), m = read(); ans = big0 = newBign(0); big2 = newBign(2); for(int t = 1; t <= n; t++){ ret = big0; for(int i = 1; i <= m; i++) pow2[i][0] = newBign(read()); initPow(); for(int i = 0; i <= m; i++) for(int j = 0; j <= m; j++) f[i][j] = big0; for(int i = 0; i <= m; i++) for(int j = 0; i + j <= m; j++){ if(i > 0) f[i][j] = max(f[i - 1][j] + pow2[i][i + j], f[i][j]); if(j > 0) f[i][j] = max(f[i][j - 1] + pow2[m - j + 1][i + j], f[i][j]); } for(int i = 0; i <= m; i++) ret = max(ret, f[i][m - i]); ans = ans + ret; } ans.print(); return 0; }
怎么说呢
#include<iostream> #include<cstdio> using namespace std; const int N = 1e4 + 5; int n, x1[N], y1[N], l[N], w[N], x, y, ans = -1; int main(){ scanf("%d", &n); for(int i = 1; i <= n; i++) scanf("%d%d%d%d", &x1[i], &y1[i], &l[i], &w[i]); scanf("%d%d", &x, &y); for(int i = n; i >= 1; i--){ if(x1[i] <= x && x <= x1[i] + l[i] - 1 && y1[i] <= y && y <= y1[i] + w[i] - 1){ ans = i; break; } } printf("%d", ans); return 0; }
标签：取数游戏 技术分享 oid noip play getchar log scan namespace
原文地址：http://www.cnblogs.com/CzYoL/p/7420565.html
