Codeforces Round #517 (Div. 2)

A

#include<queue>
#include<cstdio>
#include<cstring>
#include<algorithm>
#define rep(a,b,c) for(int a = b; a <= c;++ a)
#define per(a,b,c) for(int a = b; a >= c; -- a)
#define gc getchar()
#define pc putchar
inline int read() {
	int x = 0,f = 1;
	char c = gc;
	while(c < '0' || c >'9'){ if(c == '-') f = -1; c = gc; }
	while(c <= '9' && c >= '0') x = x * 10 + c - '0',c = gc;
	return x * f;
}
void print(int x) {
	if(x < 0) {
		pc('-'); x = -x;
	}
	if(x >= 10) print(x / 10);
	pc(x % 10 + '0') ;
} 

int main() {
	int n = read(),m =read(),k = read();
	int xu=1,xd=n,yu=1,yd=m,ans=0;
	while(k --)  {
		if(xu>xd || yu > yd) break;
		ans += 2 * (xd - xu + 1) + 2 * (yd - yu + 1) - 4;
		xd -=2 , xu += 2, yd -= 2, yu += 2;
	}
	print(ans);
	return 0;
}

B

枚举尾项往前推

#include<queue>
#include<cstdio>
#include<cstring>
#include<algorithm>
#define rep(a,b,c) for(int a = b; a <= c;++ a)
#define per(a,b,c) for(int a = b; a >= c; -- a)
#define gc getchar()
#define pc putchar
inline int read() {
	int x = 0,f = 1;
	char c = gc;
	while(c < '0' || c >'9'){ if(c == '-') f = -1; c = gc; }
	while(c <= '9' && c >= '0') x = x * 10 + c - '0',c = gc;
	return x * f;
}
void print(int x) {
	if(x < 0) {
		pc('-'); x = -x;
	}
	if(x >= 10) print(x / 10);
	pc(x % 10 + '0') ;
}
int n,k;
const int maxn = 100007;
int a[maxn][2],b[maxn][2],ans[maxn][2];
bool Check() {
	for(int i = n - 1;i;-- i)
		for(int j = 0;j < 2;++ j) {
			if(a[i][j] && b[i][j]) {
				if(!ans[i + 1][j]) return false;
				ans[i][j] = 1;
			}
			else if(!a[i][j] && !b[i][j]) {
				if(ans[i + 1][j]) return false ;
				ans[i][j] = 0;
			}
			else if(!a[i][j] && b[i][j]) return false;
			else if(a[i][j] && !b[i][j])  {
				if(ans[i + 1][j]) ans[i][j] = 0;
 				else ans[i][j] = 1;
			}
		}
	for(int i = 1;i < n;++ i)
		for(int j = 0;j < 2;++ j)
			if((a[i][j] != (ans[i][j] | ans[i + 1][j])) || (b[i][j] != (ans[i][j] & ans[i + 1][j])))
				return false;
	puts("YES");
	for(int i = 1;i <= n;++ i) print(ans[i][1] * 2 + ans[i][0]),pc(' ');
	return true;
}

int main() {
	n=read();
	for(int i =1,ai;i < n;++ i) {
		ai = read();
		a[i][0] = ai & 1;
		a[i][1] = ai >> 1 & 1;
	}
	for(int i=1,bi;i < n;++ i)  {
		bi = read();
		 b[i][0]=bi&1;
		 b[i][1] = bi >> 1 & 1;
	} 

	ans[n][0] = 0, ans[n][1] = 0;
	if(Check())return 0;
	ans[n][0] = 0, ans[n][1] = 1;
	if(Check())return 0;
	ans[n][0] = 1, ans[n][1] = 0;
	if(Check())return 0;
	ans[n][0] = 1, ans[n][1] = 1;
	if(Check())return 0;
	puts("NO"); 

	return 0;
}

C

二分最大能构成的1-n

#include<queue>
#include<cstdio>
#include<cstring>
#include<algorithm>
#define rep(a,b,c) for(int a = b; a <= c;++ a)
#define per(a,b,c) for(int a = b; a >= c; -- a)
#define gc getchar()
#define pc putchar
inline int read() {
	int x = 0,f = 1;
	char c = gc;
	while(c < '0' || c >'9'){ if(c == '-') f = -1; c = gc; }
	while(c <= '9' && c >= '0') x = x * 10 + c - '0',c = gc;
	return x * f;
}
void print(int x) {
	if(x < 0) {
		pc('-'); x = -x;
	}
	if(x >= 10) print(x / 10);
	pc(x % 10 + '0') ;
}
int tot = 0;
const int maxn = 10000007;
int ans1[maxn],a1,ans2[maxn],a2,vis[maxn];
int A,B,n,m;
bool judge(int x)  {
	int a = A,b = B;
	while(x) {
		if(a < b) std::swap(a,b);
		if(a < x) return false;
		a -= x, -- x;
	}
	return true;
}

int main() {
	A = read(),B = read(),n = 0,m = 0;
	int l=0,r=1e6,mid,ans=0;
	while(l <= r) {
		if(judge(mid = l + r >> 1)) l = mid + 1,ans = mid;
		else r = mid - 1;
	}
	for(int i = ans;i; -- i)
 		if(A < B) B -= i, ans2[++ m] = i;
 		else A -= i, ans1[++ n] = i;
	print(n);
	pc('\n');
	for(int i = n;i;-- i) print(ans1[i]),pc(' '); pc('\n');
	print(m);
	pc('\n');
	for(int i=m; i; --i) print(ans2[i]),pc(' '); pc('\n');

	return 0;
}

D

找到最远的能使得前缀为a的点，bfs求字典序最小路径

#include<vector>
#include<queue>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
#define rep(a,b,c) for(int a = b; a <= c;++ a)
#define per(a,b,c) for(int a = b; a >= c; -- a)
#define gc getchar()
#define pc putchar
#define pr pair<int,int>
inline int read() {
	int x = 0,f = 1;
	char c = gc;
	while(c < '0' || c >'9'){ if(c == '-') f = -1; c = gc; }
	while(c <= '9' && c >= '0') x = x * 10 + c - '0',c = gc;
	return x * f;
}
void print(int x) {
	if(x < 0) {
		pc('-'); x = -x;
	}
	if(x >= 10) print(x / 10);
	pc(x % 10 + '0') ;
}
const int maxn = 4007;
#define mp make_pair
#define fi first
#define se second
int n,k,f[maxn][maxn],mx;
char s[maxn][maxn];
vector<pr> v;
#define INF 0x3f3f3f3f;
bool vis[maxn][maxn];
void bfs() {
    vector<int> path;
    memset(vis,0,sizeof(vis));
    vector<pr> nxt;
    int xxx = v[1].fi,yyy = v[1].se;
    for(int i = 0;i < v.size();++ i) {
		int X = v[i].fi,Y = v[i].se;
		vis[X][Y] = 1;
		nxt.push_back(v[i]);
	}
    for(int i = 0; i < n * 2 - (xxx + yyy); i++) {
        int min_col = INF;
        for(int j = 0;j < nxt.size(); j++){
            int xx = nxt[j].fi,yy = nxt[j].se;
            for(int k = 0;k <= 1;++ k) {
				int tx = (k & 1) ? xx + 1 : xx;
				int ty = !(k & 1) ? yy + 1 : yy;
				if(tx > n || ty > n) continue;
                min_col = min(min_col,s[tx][ty] - 'a');
            }
        }
        path.push_back(min_col);
        vector<pr> nxt2;
        for(int j = 0; j < nxt.size(); j++) {
            int xx = nxt[j].fi,yy = nxt[j].se;
            for(int k = 0;k <= 1;++ k) {
                int tx = (k & 1) ? xx + 1 : xx;
				int ty = !(k & 1) ? yy + 1 : yy;
				if(tx > n || ty > n) continue;
                if(!vis[tx][ty] && s[tx][ty] - 'a' == min_col) {
                    vis[tx][ty] = 1;
                    nxt2.push_back(mp(tx,ty));
                }
            }
        }
        nxt = nxt2;
    }
    for(int i = 1; i <= mx; i++) pc('a');
	for(int i = 0; i < path.size(); i ++) {
            pc(path[i] + 'a');
    }
}
int GG,id[maxn][maxn];
int main() {
	n = read(); k = read();
	for(int i = 1; i <= n; i++) scanf("%s", s[i] + 1);
	for(int i = 1; i <= n; i++) s[i][0] = 'z' + 1;
	for(int i = 1; i <= n; i++) s[0][i] = s[n + 1][i] = 'z' + 1;
	for(int i = 1; i <= n; i++) {
		for(int j = 1; j <= n; j++) {
			id[i][j] = ++GG;
			f[i][j] = max(f[i - 1][j], f[i][j - 1]);
			if(s[i][j] == 'a') f[i][j]++;
		}
	}
	for(int i = 1; i <= n; i++) {
		for(int j = 1; j <= n; j++) {
			int ned = i + j - f[i][j] - 1;
			if(ned <= k) {
				if(i + j > mx) v.clear(), v.push_back(mp(i, j)), mx = i + j;
				else if(i + j == mx) v.push_back(mp(i, j));
			}
		}
	}
	mx --;
	//print(mx);
	bfs();
    return 0;
}