Codeforces Good Bye 2016（部分题解）
Posted AC_Arthur
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本次比赛一共AC了前4题...
A. New Year and Hurry
水题。
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <iostream>
#include <string>
#include <vector>
#include <stack>
#include <ctime>
#include <bitset>
#include <cstdlib>
#include <cmath>
#include <set>
#include <list>
#include <deque>
#include <map>
#include <queue>
#define Max(a,b) ((a)>(b)?(a):(b))
#define Min(a,b) ((a)<(b)?(a):(b))
using namespace std;
typedef long long ll;
typedef long double ld;
const double eps = 1e-6;
const double PI = acos(-1);
const int mod = 1000000000 + 7;
const int INF = 0x3f3f3f3f;
// & 0x7FFFFFFF
const int seed = 131;
const ll INF64 = ll(1e18);
const int maxn = 1e5 + 10;
int k, n;
int main()
scanf("%d%d", &n, &k);
int sum = 0, ans = 0;
for(int i = 1; i <= n; i++)
sum += 5*i;
if(sum + k <= 4*60) ++ans;
printf("%d\\n", ans);
return 0;
B. New Year and North Pole
其实向东西跑我们不用理睬（在南北极除外）， 这样只需要记录当前距离北极多远就行了。
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <iostream>
#include <string>
#include <vector>
#include <stack>
#include <ctime>
#include <bitset>
#include <cstdlib>
#include <cmath>
#include <set>
#include <list>
#include <deque>
#include <map>
#include <queue>
#define Max(a,b) ((a)>(b)?(a):(b))
#define Min(a,b) ((a)<(b)?(a):(b))
using namespace std;
typedef long long ll;
typedef long double ld;
const double eps = 1e-6;
const double PI = acos(-1);
const int mod = 1000000000 + 7;
const int INF = 0x3f3f3f3f;
// & 0x7FFFFFFF
const int seed = 131;
const ll INF64 = ll(1e18);
const int maxn = 1e5 + 10;
int k, n;
char s[1000];
int main()
scanf("%d", &n);
int dist = 0;
bool ok = true;
for(int i = 1; i <= n; i++)
int v; scanf("%d%s", &v, s);
if(s[0] == 'E' || s[0] == 'W')
if(dist == 0 || dist == 20000) ok = false;
else if(s[0] == 'N')
if(dist - v < 0) ok = false;
else dist -= v;
else
if(dist + v > 20000) ok = false;
else dist += v;
if(ok && dist == 0) printf("YES\\n");
else printf("NO\\n");
return 0;
C. New Year and Rating
我们假设初始可能值为x， 那么对于前i次的累加值cur， 第i+1次如果是div1， 那么x+cur >= 1900， 否则， x+cur < 1900， 这是一个不等式， 求出最终范围就行了。
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <iostream>
#include <string>
#include <vector>
#include <stack>
#include <ctime>
#include <bitset>
#include <cstdlib>
#include <cmath>
#include <set>
#include <list>
#include <deque>
#include <map>
#include <queue>
#define Max(a,b) ((a)>(b)?(a):(b))
#define Min(a,b) ((a)<(b)?(a):(b))
using namespace std;
typedef long long ll;
typedef long double ld;
const double eps = 1e-6;
const double PI = acos(-1);
const int mod = 1000000000 + 7;
const int INF = 0x3f3f3f3f;
// & 0x7FFFFFFF
const int seed = 131;
const ll INF64 = ll(1e18);
const int maxn = 2e5 + 10;
int k, n, a[maxn], b[maxn];
bool ok(int mid)
int ans = mid;
for(int i = 1; i <= n; i++)
if(b[i] == 2)
if(ans >= 1900) return false;
ans += a[i];
return true;
bool hehe(int mid)
int cur = mid;
for(int i = 1; i <= n; i++)
if(b[i] == 1)
if(cur < 1900) return false;
else
if(cur >= 1900) return false;
cur += a[i];
return true;
char s[1000];
int main()
scanf("%d", &n);
for(int i = 1; i <= n; i++)
scanf("%d%d", &a[i], &b[i]);
int res = 0, maxv = INF, minv = -INF;
for(int i = 1; i <= n; i++)
if(b[i] == 1)
minv = max(minv, 1900-res);
else
maxv = min(maxv, 1899-res);
res += a[i];
// printf("%d %d-\\n", minv, maxv);
if(minv > maxv) printf("Impossible\\n");
else if(maxv == INF) printf("Infinity\\n");
else
for(int i = 1; i <= n; i++)
maxv += a[i];
printf("%d\\n", maxv);
return 0;
D. New Year and Fireworks
我们用dp[ceng][t][x][y][dir]表示状态：在(x, y)位置方向为dir，处于第ceng层烟花的第t个时间。 如果这个状态访问过剪枝即可。
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <iostream>
#include <string>
#include <vector>
#include <stack>
#include <ctime>
#include <bitset>
#include <cstdlib>
#include <cmath>
#include <set>
#include <list>
#include <deque>
#include <map>
#include <queue>
#define Max(a,b) ((a)>(b)?(a):(b))
#define Min(a,b) ((a)<(b)?(a):(b))
using namespace std;
typedef long long ll;
typedef long double ld;
const double eps = 1e-6;
const double PI = acos(-1);
const int mod = 1000000000 + 7;
const int INF = 0x3f3f3f3f;
// & 0x7FFFFFFF
const int seed = 131;
const ll INF64 = ll(1e18);
const int maxn = 150*2 + 10;
int n, res = 0, t[maxn], vis[maxn][maxn];
bool d[30][6][maxn][maxn][9];
void dp(int ceng, int tt, int x, int y, int dir)
bool& ans = d[ceng][tt][x][y][dir];
if(ceng >= n) return ;
if(ans) return ;
ans = 1;
if(!vis[x][y]) res++, vis[x][y] = 1;
if(dir == 0)
if(tt == t[ceng]) dp(ceng+1, 1, x-1, y+1, 1), dp(ceng+1, 1, x+1, y+1, 7);
else dp(ceng, tt+1, x, y+1, dir);
else if(dir == 1)
if(tt == t[ceng]) dp(ceng+1, 1, x, y+1, 0), dp(ceng+1, 1, x-1, y, 2);
else dp(ceng, tt+1, x-1, y+1, dir);
else if(dir == 2)
if(tt == t[ceng]) dp(ceng+1, 1, x-1, y+1, 1), dp(ceng+1, 1, x-1, y-1, 3);
else dp(ceng, tt+1, x-1, y, dir);
else if(dir == 3)
if(tt == t[ceng]) dp(ceng+1, 1, x-1, y, 2), dp(ceng+1, 1, x, y-1, 4);
else dp(ceng, tt+1, x-1, y-1, dir);
else if(dir == 4)
if(tt == t[ceng]) dp(ceng+1, 1, x-1, y-1, 3), dp(ceng+1, 1, x+1, y-1, 5);
else dp(ceng, tt+1, x, y-1, dir);
else if(dir == 5)
if(tt == t[ceng]) dp(ceng+1, 1, x, y-1, 4), dp(ceng+1, 1, x+1, y, 6);
else dp(ceng, tt+1, x+1, y-1, dir);
else if(dir == 6)
if(tt == t[ceng]) dp(ceng+1, 1, x+1, y-1, 5), dp(ceng+1, 1, x+1, y+1, 7);
else dp(ceng, tt+1, x+1, y, dir);
else if(dir == 7)
if(tt == t[ceng]) dp(ceng+1, 1, x, y+1, 0), dp(ceng+1, 1, x+1, y, 6);
else dp(ceng, tt+1, x+1, y+1, dir);
int main()
scanf("%d", &n);
for(int i = 0; i < n; i++)
scanf("%d", &t[i]);
dp(0, 1, 160, 160, 0);
printf("%d\\n", res);
return 0;
就会这几道水题， 我也差不多是个废人了。
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