u1s1,zyyyyy出的题目真的好好，补题的过程学到了好多（好菜啊o(╥﹏╥)o
题目链接
只贴了代码，题解参考此处
A
签到
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#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
int main(){
int n;
int a[105];
cin >> n;
int s, m, b;
s = m = b = 0;
for(int i = 0; i < n; ++i){
double x;
cin >> x;
if(x == 2 or x == 2.5 or x == 3) s++;
else if(x == 5 or x == 5.5 or x == 6) m++;
else b++;
}
cout << "Small: " << s << endl << "Medium: " << m << endl << "Big: " << b << endl;
return 0;
}
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B
贪心
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#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
ll a[100005];
ll b[100005];
int main(){
ll n;
ll x, y, z;
cin >> n;
for(int i = 1; i <= n; ++i) cin >> a[i];
cin >> x >> y >> z;
ll mi = 100000000000;
ll ans = n;
b[0] = x;
for(int i = 1; i <= n; ++i) b[i] = b[i - 1] + z;
bool ok = true;
for(int i = n; i > 0; --i){
if(b[i] <= a[i]){
ok = false;
break;
}
if(b[i] - z - y > a[i] and mi > (y + z)){
ans--;
b[i] = b[i] - z - y;
mi = min(b[i] - a[i], mi- y - z);
}
mi = min(mi, b[i] - a[i]);
}
if(!ok) puts("HHUTQL");
else cout << ans << endl;
return 0;
}
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C
线段树
没板子就不写了 o(╥﹏╥)o
D
状态压缩 / dfs暴搜
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#include "bits/stdc++.h"
using namespace std;
typedef long long ll;
ll n, m, c;
ll w[25];
ll v[25];
vector<pair<ll, ll>> conflict;
ll ind[1005];
ll to_ind(ll x){
return ind[x];
}
bool ok(ll x){
for(auto i : conflict){
ll l = i.first;
ll r = i.second;
if((x >> l & 1) and (x >> r & 1)) return false;
}
return true;
}
ll cal(ll x){
ll sumw = 0;
ll sumv = 0;
for(ll i = 0; i < n; ++i){
if(x >> i & 1){
sumw += w[i];
sumv += v[i];
if(sumw > m) return 0;
}
}
return sumv;
}
int main() {
ll _;
cin >> _;
while (_--) {
conflict.clear();
cin >> n >> m >> c;
for(ll i = 0; i < n; ++i){
ll x, y, z;
cin >> x >> y >> z;
w[i] = y;
v[i] = z;
ind[x] = i;
}
for(ll i = 0; i < c; ++i){
ll x, y;
cin >> x >> y;
conflict.emplace_back(to_ind(x), to_ind(y));
}
ll ans = 0;
for(ll i = 0; i < (1 << n); ++i){
if(ok(i)){
ans = max(ans, cal(i));
}
}
cout << ans << endl;
}
return 0;
}
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E
打表 矩阵快速幂
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#include "bits/stdc++.h"
using namespace std;
using ll = long long;
const ll mod = 1000000007;
typedef vector<ll> vec;
typedef vector<vector<ll>> mat;
mat mul(mat &A , mat &B){
mat C(A.size() , vec(B[0].size()));
for(ll i = 0; i < A.size() ; ++i){
for(ll k = 0 ; k < B.size() ; ++k){
for(ll j = 0 ; j < B[0].size() ; ++j){
C[i][j] = (C[i][j] + A[i][k] * B[k][j]) % mod;
}
}
}
return C;
}
mat pow_mat(mat A , ll n){
mat B(A.size() , vec(A.size()));
for(ll i = 0 ; i < A.size() ; ++i){
B[i][i] = 1;
}
while(n > 0){
if(n & 1) B = mul(B , A);
A = mul(A , A);
n >>= 1;
}
return B;
}
int main() {
// ll n = 1;
// ll ans = 0;
// vector<ll> vt;
// vector<ll> tt;
// for(ll i = 0; i < n; ++i){
// tt.push_back(i % 3);
// }
// for(ll i = 0; i < (1 << n); ++i){
// vt.clear();
// for(ll j = 0; j < n; ++j){
// if(i >> j & 1){
// vt.push_back(j % 3);
// }
// }
// bool ok = true;
// for(ll j = 0; j < vt.size(); ++j){
// if(vt[j] != tt[j]){
// ok = false;
// break;
// }
// }
// if(ok) ans++;
// }
// cout << ans - 1 << endl;
ll g1 = 1, g2 = 2, g3 = 3;
ll _;
cin >> _;
while (_--) {
//此处有md渲染有bug，所以这样写
mat core = {{1, 0, 1, 1}}, {1, 0, 0, 0}, {0, 1, 0, 0}, {{0, 0, 0, 1}};
ll n;
cin >> n;
if(n <= 3){
cout << n << endl;
continue;
}
n -= 3;
core = pow_mat(core, n);
ll ans = 0;
ans += core[0][0] * g3;
ans %= mod;
ans += core[0][1] * g2;
ans %= mod;
ans += core[0][2] * g1;
ans %= mod;
ans += core[0][3];
ans %= mod;
cout << ans << endl;
}
return 0;
}
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F
分治 01背包 二分 前缀
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#include "bits/stdc++.h"
using namespace std;
typedef long long ll ;
int n, c;
const int maxn = 1005;
const int maxm = 32007;
int dp1[maxm];
int dp2[maxm];
int n1;
int n2;
int W;
vector<int> w1, v1, w2, v2;
const int limit = (1 << 10);
vector<pair<int, int>> vt1;
vector<int> vt2;
int premax[maxm];
// 1 for big , 2 for small
int main() {
cin >> n >> c;
memset(premax, -1, sizeof(premax));
W = maxm - 3;
// for test
w1.push_back(0);
w2.push_back(0);
v1.push_back(0);
v2.push_back(0);
for(int i = 0; i < n; ++i){
int x, y;
cin >> x >> y;
if(x >= limit){
++n1;
w1.push_back((x >> 10));
v1.push_back(y);
}else {
++n2;
w2.push_back(x);
v2.push_back(y);
}
}
for(int i = 1; i <= n1; ++i)
for(int j = W; j >= w1[i]; --j)
dp1[j] = max(dp1[j], dp1[j - w1[i]] + v1[i]);
for(int i = 1; i <= n2; ++i)
for(int j = W; j >= w2[i]; --j)
dp2[j] = max(dp2[j], dp2[j - w2[i]] + v2[i]);
for(int i = 0; i <= W; ++i){
vt1.emplace_back((i << 10), dp1[i]);
}
premax[0] = dp2[0];
for(int i = 1; i <= W; ++i){
premax[i] = max(premax[i - 1], dp2[i]);
}
int ans = 0;
for(int i = 0; i < W; ++i){
int cnt = vt1[i].first;
int rem = c - cnt;
if(rem >= 0 and rem < W) ans = max(ans, vt1[i].second + premax[rem]);
}
cout << ans << endl;
return 0;
}
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G
图论 dfs
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#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
int V;
const int maxv = 1e5 + 10;
vector<int> G[maxv];
int od[maxv];
bool ok ;
void dfs(int x, int f, int d){
for(int i = 0; i < G[x].size(); ++i){
int to = G[x][i];
if(to == f) continue;
if(od[to] != 0){
if((d + 1 - od[to]) & 1){
ok = true;
break;
}
}else {
od[to] = d + 1;
dfs(to, x, d + 1);
}
}
}
int main(){
int m;
cin >> V >> m;
for(int i = 0; i < m; ++i){
int u, v;
cin >> u >> v;
G[u].push_back(v);
G[v].push_back(u);
}
if(m >= V){
for(int i = 1; i <= V; ++i){
if(od[i] == 0){
dfs(i, -1, 1);
od[i] = 1;
}
}
}else ok = false;
if(ok) puts("yes"); else puts("NO");
return 0;
}
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H
欧拉降幂 快速幂 快速乘
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#include "bits/stdc++.h"
using namespace std;
typedef long long ll ;
inline ll qmul(ll x, ll y, ll mod)
{
return ( x * y - (ll) ( (long double) x / mod*y )*mod + mod ) % mod;
}
ll qpow(ll x, ll n, ll mod){
ll res =1;
while(n>0)
{
if(n&1) res=qmul(res, x, mod)%mod;
x=qmul(x, x, mod)%mod;
n>>=1;
}
return res;
}
ll phi(ll n)
{
ll res=n;
for(ll i=2;i*i<=n;i++)
{
if(n%i==0)
{
res=res/i*(i-1);
while(n%i==0) n/=i;
}
}
if(n!=1) res=res/n*(n-1);
return res;
}
char n[100005];
int main() {
int _;
cin >> _;
while (_--) {
ll ans ;
ll m;
scanf("%s", n);
scanf("%lld", &m);
ll tmpn = 0;
ll tmp_phi = phi(m);
ll len = strlen(n);
if(len <= 16){
ll tmp_mul = 1;
for(int i = len - 1; i >= 0; --i){
tmpn += (n[i] - '0') * tmp_mul;
tmp_mul *= 10;
}
ll tmp_ans = qpow(3, tmpn, m);
ans = tmp_ans;
ans -= 2;
ans += m;
ans %= m;
}else {
for(int i = 0; i < len; ++i){
tmpn = tmpn * 10 + (n[i] - '0');
tmpn %= tmp_phi;
}
tmpn += tmp_phi;
ll tmp_ans = qpow(3, tmpn, m);
ans = tmp_ans;
ans -= 2;
ans += m;
ans %= m;
}
cout << ans << endl;
}
return 0;
}
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I
二分 贪心
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#include "bits/stdc++.h"
using namespace std;
typedef long long ll ;
const int maxn = 1e5 + 10;
int a[maxn];
int b[maxn];
int n, m;
bool check(int x){
int i = 0, j = 0;
while(i < n and j < m){
if(abs(a[i] - b[j]) <= x){
++i, ++j;
}else {
++j;
}
}
return i == n;
}
int main() {
ios::sync_with_stdio(false);
cin.tie(0);
cout.tie(0);
int _;
cin >> _;
while (_--) {
cin >> n >> m;
for(int i = 0; i < n; ++i){
cin >> a[i];
}
for(int i = 0; i < m; ++i){
cin >> b[i];
}
sort(a, a + n);
sort(b, b + m);
int l = 0, r = 5 * 1e8 + 3;
int mid;
while (l <= r){
mid = (l + r) >> 1;
if(check(mid)){
r = mid - 1;
}else {
l = mid + 1;
}
}
cout << l << endl;
}
return 0;
}
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