AtCoder Beginner Contest 317(D-G)
D - President (atcoder.jp)
（1）题目大意
（2）解题思路
考虑到z最大不超过1e5，N最多不超过100，因此可以考虑用背包来写，dp[j]表示拿高桥拿j分最少需要花费多少个选民转换，最后把答案取个min即可。
（3）代码实现
#include
#define rep(i,z,n) for(int i = z;i <= n; i++)
#define per(i,n,z) for(int i = n;i >= z; i--)
#define PII pair
#define fi first
#define se second
#define vi vector
#define vl vector
#define pb push_back
#define sz(x) (int)x.size()
#define all(x) (x).begin(),(x).end()
using namespace std;
using ll = long long;
const int N = 2e5 + 10;
PII lose[N];
int cnt = 0;
ll dp[N];
const ll inf = 0x3f3f3f3f3f3f3f3f;
void solve()
{
int n;
cin >> n;
int win = 0,tot = 0;
for(int i = 1;i <= n;i ++) {
int x,y,w;
cin >> x >> y >> w;
tot += w;
if(x > y) win += w;
else lose[++ cnt] = {(x + y + 1) / 2 - x,w};
}
if(2 * win > tot) {
cout << 0 << endl;
return;
}
memset(dp,0x3f,sizeof(dp));
dp[0] = 0;
for(int i = 1;i <= cnt;i ++) {
for(int j = tot;j >= lose[i].se;j --) {
if(dp[j - lose[i].se] != inf) dp[j] = min(dp[j],dp[j - lose[i].se] + lose[i].fi);
}
}
int has = (tot + 1) / 2 - win;
ll ok = inf;
for(int i = has;i <= tot;i ++) {
ok = min(ok,dp[i]);
}
cout << ok << endl;
}
int main()
{
ios::sync_with_stdio(false);
cin.tie(0),cout.tie(0);
int T = 1;
// cin >> T;
while(T --) solve();
return 0;
} E - Avoid Eye Contact (atcoder.jp)
（1）题目大意
（2）解题思路
考虑预处理出所有障碍物点，或者能被看到得点，跑一遍bfs即可。
（3）代码实现
#include
#define rep(i,z,n) for(int i = z;i <= n; i++)
#define per(i,n,z) for(int i = n;i >= z; i--)
#define PII pair
#define fi first
#define se second
#define vi vector
#define vl vector
#define pb push_back
#define sz(x) (int)x.size()
#define all(x) (x).begin(),(x).end()
using namespace std;
using ll = long long;
const int N = 2e3 + 10;
int L[N][N],R[N][N],U[N][N],D[N][N];
bool obs[N][N];
int dis[N][N];
string s[N];
const int inf = 0x3f3f3f3f;
int mx[4] = {1,-1,0,0};
int my[4] = {0,0,1,-1};
void solve()
{
int n,m;
cin >> n >> m;
for(int i = 1;i <= n;i ++) {
cin >> s[i];
s[i] = " " + s[i];
}
for(int i = 1;i <= n;i ++) {
for(int j = 1;j <= m;j ++) {
if(s[i][j] == '>') {
R[i][j] = true;
}
else if(s[i][j] == '.') R[i][j] |= R[i][j - 1];
}
for(int j = m;j >= 1;j --) {
if(s[i][j] == '<') {
L[i][j] = true;
}
else if(s[i][j] == '.') L[i][j] |= L[i][j + 1];
}
}
for(int i = 1;i <= n;i ++) {
for(int j = 1;j <= m;j ++) {
if(obs[i][j]) cout << i << ' ' << j << endl;
}
}
for(int i = 1;i <= m;i ++) {
for(int j = 1;j <= n;j ++) {
if(s[j][i] == 'v') {
D[j][i] = true;
}
else if(s[j][i] == '.') D[j][i] |= D[j - 1][i];
}
for(int j = n;j >= 1;j --) {
if(s[j][i] == '^') {
U[j][i] = true;
}
else if(s[j][i] == '.') U[j][i] |= U[j + 1][i];
}
}
int sx,sy;
int ex,ey;
for(int i = 1;i <= n;i ++) {
for(int j = 1;j <= m;j ++) {
if(s[i][j] == '#') obs[i][j] = true;
obs[i][j] |= L[i][j] | R[i][j] | U[i][j] | D[i][j];
if(s[i][j] == 'S') sx = i,sy = j;
if(s[i][j] == 'G') ex = i,ey = j;
}
}
// for(int i = 1;i <= n;i ++) {
// for(int j = 1;j <= m;j ++) {
// if(obs[i][j]) cout << i << ' ' << j << endl;
// }
// }
auto bfs = [&](int sx,int sy) {
queue q;
q.push({sx,sy});
memset(dis,0x3f,sizeof(dis));
dis[sx][sy] = 0;
while(!q.empty()) {
auto [x,y] = q.front();
q.pop();
for(int i = 0;i < 4;i ++) {
int dx = mx[i] + x,dy = my[i] + y;
if(dx <= 0 || dy <= 0 || dx > n || dy > m || obs[dx][dy]) continue;
if(dis[dx][dy] > dis[x][y] + 1) {
dis[dx][dy] = dis[x][y] + 1;
q.push({dx,dy});
}
}
}
};
bfs(sx,sy);
if(dis[ex][ey] == inf) cout << -1 << endl;
else cout << dis[ex][ey] << endl;
}
int main()
{
ios::sync_with_stdio(false);
cin.tie(0),cout.tie(0);
int T = 1;
// cin >> T;
while(T --) solve();
return 0;
} F - Nim (atcoder.jp)
（1）题目大意
（2）解题思路
考虑数字太大，我们用数位dp计数，状态为dp[pos][i][j][k][r1][r2][r3]，表示在二进制位为pos这位时，我们第一个数字填i，第二个数字填j，第三个数字填k，并且三个数得余数分别为r1，r2，r3得方案数有多少。
显然对于非法情况若某一位i^j^k!=0，则不行，因为在异或操作下，若这一位异或不为零了，那么就永远不可能为0。
对于答案非法情况，首先全0得方案需要减去，其次是有两个数相同，一个数为0得情况也要减去。
（3）代码实现
#include
#define rep(i,z,n) for(int i = z;i <= n; i++)
#define per(i,n,z) for(int i = n;i >= z; i--)
#define PII pair
#define fi first
#define se second
#define vi vector
#define vl vector
#define pb push_back
#define sz(x) (int)x.size()
#define all(x) (x).begin(),(x).end()
using namespace std;
using ll = long long;
const int N = 61;
const ll mod = 998244353;
ll dp[N][2][2][2][2][2][2][10][10][10];
vector nums;
int a,b,c;
ll ksm(ll a,ll b)
{
ll rs = 1;
while(b) {
if(b & 1) rs = rs * a % mod;
b >>= 1;
a = a * a % mod;
}
return rs;
}
ll calc(int pos,bool z1,bool z2,bool z3,int pi,int pj,int pk,int r1,int r2,int r3)
{
if(pos == -1) return (!r1 && !r2 && !r3);
if(dp[pos][z1][z2][z3][pi][pj][pk][r1][r2][r3] != -1) return dp[pos][z1][z2][z3][pi][pj][pk][r1][r2][r3];
ll res = 0;
int up1 = z1 ? nums[pos] : 1;
int up2 = z2 ? nums[pos] : 1;
int up3 = z3 ? nums[pos] : 1;
for(int i = 0;i <= up1;i ++) {
for(int j = 0;j <= up2;j ++) {
for(int k = 0;k <= up3;k ++) {
if(i ^ j ^ k) continue;
int nr1 = (i == 1) ? (r1 + (1ll << pos)) % a : r1;
int nr2 = (j == 1) ? (r2 + (1ll << pos)) % b : r2;
int nr3 = (k == 1) ? (r3 + (1ll << pos)) % c : r3;
res += calc(pos - 1,z1 && i == up1,z2 && j == up2,z3 && k == up3,i,j,k,nr1,nr2,nr3);
res %= mod;
// cout << i << ' ' << j << ' ' << k << ' ' << res << endl;
}
}
}
return dp[pos][z1][z2][z3][pi][pj][pk][r1][r2][r3] = res;
}
void solve()
{
ll n;
cin >> n >> a >> b >> c;
ll rn = n;
while(n) {
nums.pb(n % 2);
n /= 2;
}
memset(dp,-1,sizeof(dp));
calc(sz(nums) - 1,true,true,true,0,0,0,0,0,0);
ll ans = 0;
for(int i = 0;i <= 1;i ++) {
for(int j = 0;j <= 1;j ++) {
for(int k = 0;k <= 1;k ++) {
for(int z1 = 0;z1 <= 1;z1 ++) {
for(int z2 = 0;z2 <= 1;z2 ++) {
for(int z3 = 0;z3 <= 1;z3 ++) {
if(dp[sz(nums) - 1][z1][z2][z3][i][j][k][0][0][0] != -1) ans += dp[sz(nums) - 1][z1][z2][z3][i][j][k][0][0][0];
ans %= mod;
}
}
}
}
}
}
n = rn;
ans -= 1;
ans -= n / lcm(a,b);
ans -= n / lcm(a,c);
ans -= n / lcm(b,c);
ans = (ans % mod + mod) % mod;
cout << ans << endl;
}
int main()
{
ios::sync_with_stdio(false);
cin.tie(0),cout.tie(0);
int T = 1;
// cin >> T;
while(T --) solve();
return 0;
} G - Rearranging (atcoder.jp)
（1）题目大意
（2）解题思路
我们发现这个题很有可能是全部可以得，（事实上确实全部可以，由Hall定理可知），那么考虑怎么构造方案，由Hall定理得，我们得最大完美匹配在删除一些匹配后依旧成立，因此我们只需要一列一列得匹配完即可。
（3）代码实现
#include
#define rep(i,z,n) for(int i = z;i <= n; i++)
#define per(i,n,z) for(int i = n;i >= z; i--)
#define PII pair
#define fi first
#define se second
#define vi vector
#define vl vector
#define pb push_back
#define sz(x) (int)x.size()
#define all(x) (x).begin(),(x).end()
using namespace std;
using ll = long long;
const int N = 105;
bool vis[N];
vector v[N];
int match[N],Ans[N][N];
bool find(int x)
{
vis[x] = true;
for(auto y : v[x]) {
if(!match[y] || (!vis[match[y]] && find(match[y]))) {
match[y] = x;
return true;
}
}
return false;
}
void solve()
{
int n,m;
cin >> n >> m;
rep(i,1,n) rep(j,1,m) {
int x;
cin >> x;
v[i].pb(x);
}
rep(i,1,m) {
memset(match,0,sizeof(match));
rep(j,1,n) {
memset(vis,false,sizeof(vis));
if(!find(j)) {
cout << "No" << '\n';
return;
}
}
rep(j,1,n) {
Ans[match[j]][i] = j;
v[match[j]].erase(find(v[match[j]].begin(),v[match[j]].end(),j));
}
}
cout << "Yes" << '\n';
rep(i,1,n) {
rep(j,1,m) cout << Ans[i][j] << ' ';
cout << '\n';
}
}
int main()
{
ios::sync_with_stdio(false);
cin.tie(0),cout.tie(0);
int T = 1;
// cin >> T;
while(T --) solve();
return 0;
}