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 <bits/stdc++.h>
#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<int,int>
#define fi first
#define se second
#define vi vector<int>
#define vl vector<ll>
#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 <bits/stdc++.h>
#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<int,int>
#define fi first
#define se second
#define vi vector<int>
#define vl vector<ll>
#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<PII> 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 <bits/stdc++.h>
#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<int,int>
#define fi first
#define se second
#define vi vector<int>
#define vl vector<ll>
#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<int> 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 <bits/stdc++.h>
#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<int,int>
#define fi first
#define se second
#define vi vector<int>
#define vl vector<ll>
#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<int> 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;
}
