CodeForces 550C Divisibility by Eight
这题打的时候时候打了个表，除了0、8，就没找出规律了，当时直觉是一定能通过有限的组合，在0-9之间组合出8的倍数，卡在找数的规律和如何组合数了
规律结论
只需要暴力判1-3位就行
官方题解给的是dp：https://codeforces.com/blog/entry/18329
int main() {
IOS;
// freopen("P1908_6.in","r",stdin);//读入数据
// freopen("P1908.out","w",stdout); //输出数据
string s;
cin >> s;
int len = s.size();
for (int i = 0; i < len; ++i){
if((s[i] - '0') % 8 == 0){
cout << "YES"<<endl<<s[i];
return 0;
}
for (int j = i + 1; j < len; ++j){
int temp = (s[i] - '0') * 10 + (s[j] - '0');
if(temp % 8 == 0){
//cout << temp << " ";
cout << "YES"<<endl<<s[i]<<s[j];
return 0;
}
int t = temp;
for (int k = j + 1; k < len; ++k){
temp = t;
//cout << temp << " ";
temp = temp * 10 + (s[k] - '0');
if(temp % 8 == 0){
cout << "YES"<<endl<<s[i]<<s[j]<<s[k];
return 0;
}
}
}
//cout << endl;
}
cout << "NO";
return 0;
}
CodeForces 7C Line
板子题，板子不对，sad
#define ll long long
ll exgcd(ll a, ll b, ll &x, ll &y){
if(!b){
x = 1, y = 0;
return a;
}
ll d = exgcd(b, a % b, y, x);
y -= a / b * x;
return d;
}
int main(){
ll a, b, c, x, y;
cin >> a >> b >> c;
ll d = exgcd(a, b, x, y);
if(c % d) {
cout << -1 << endl;
return 0;
}
x *= -c / d;
y *= -c / d;
cout << x << " " << y;
return 0;
}
CodeForces 977F Consecutive Subsequence
map原来还可以这么用，大震惊
因为连续上升，所以直接用map存取数量最大的值，选取最终的数量最大的值进行倒推
#define ll long long
const int maxn = 2e5 + 5;
map<ll,ll> mp;
ll a[maxn];
int main(){
ll n;
cin >> n;
ll max_id = 0, max_num = 0;
for (int i = 1; i <= n; ++i){
cin >> a[i];
mp[a[i]] = max(mp[a[i]], mp[a[i] - 1] + 1);
}
for(auto w : mp){
if(max_id < w.second){
max_num = w.first;
max_id = w.second;
}
}
vector<ll> ans;
for (int i = n; i >= 1; --i){
if(a[i] == max_num){
ans.push_back(i);
--max_num;
}
}
cout << ans.size() << endl;
reverse(ans.begin(), ans.end());
for(auto w: ans){
cout << w << " ";
}
return 0;
}
CodeForces 1066A Vova and Train
很简单的题，自己写直接交了两发wa，还是队友写的
区间问题需要加强练习，老是对区间问题头疼（对啥都头疼 ）
#define ll long long
int main(){
ll t;
cin >> t;
while(t--){
ll l, v, x, y;
cin >> l >> v >> x >> y;
if(x > y)
swap(x, y);
ll num = l / v;
ll temp = y / v - (x - 1) / v;
num -= temp;
cout << num << endl;
}
return 0;
}
CodeForces 1539A Contest Start
队友独自写的，和另一个队友推不出来，我不明白她在说什么，结果是道A…
好神奇这道题！完全想不到！证明了我确实应该多刷这种思维题
n - t / x是中间选手的位置
谢谢万能群友的解答，刻烟入肺，我还想是不是什么需要记的结论
#define ll long long
int main(){
ll k;
cin >> k;
while(k--){
ll n, x, t;
cin >> n >> x >> t;
ll m = t / x;
ll ans = 0;
if(m >= n){
ans = (n * (n - 1)) / 2;
}
else {
ans = (n - m) * m + (m * (m - 1)) / 2;
}
cout << ans << endl;
}
return 0;
}
CodeForces 1575J Jeopardy of Dropped Balls
暴力模拟
谁懂，传参把数组换成变量就能过
int n, m, k;
int g[1100][1100];
int c[1100];
int row[1100];//标记列 反向优化的纪念
// void check(){
// }
int bfs(int st){
int x = st, y = 1;//x列
while(y <= n){
if(g[y][x] == 1) {
g[y][x] = 2;
++x;
}
else if(g[y][x] == 2){
++y;
}
else if(g[y][x] == 3){
g[y][x] = 2;
--x;
}
//cout << x << " " << y << endl;
}
return x;
}
int main(){
cin >> n >> m >> k;
for (int i = 1; i <= n; ++i){
for (int j = 1; j <= m; ++j){
cin >> g[i][j];
}
}
for (int i = 1; i <= k; ++i){
int t;
cin >> t;
cout << bfs(t) << " ";
}
return 0;
}
CodeForces 1526B I Hate 1111
当时看到题目我还想是不是题目会有什么玄机，但我觉得不可能这么直白吧
也是考虑了怎么拆，wa后没思路，队友猜测是不是什么规律，被否了，真是规律
两个结论任选其一就能得出答案：
1、n*m-n-m以上的数，n、m可以任意构成
2、1111等可由11、111构成
#include<iostream>
using namespace std;
#define ll long long
int main(){
int t;
cin >> t;
while(t--){
ll n;
cin >> n;
if(n % 11 == 0 || n % 111 == 0){
cout << "YES" << endl;
continue;
}
else {
bool flag = false;
while(n >= 11){
if(n % 11 == 0) {
flag = true;
break;
}
n -= 111;
}
if(flag)
cout << "YES" << endl;
else
cout << "NO" << endl;
}
}
return 0;
}
CodeForces 1208B Uniqueness
这道题用了set wa9，重新看了题是区间，感觉没法做；下来看了题解，好像还是很难的样子…
还有一种方法，但感觉比这个难
记下最后第一个重复的数，保证区间的扩大，取最小的区间确保能被覆盖完（没法说清楚，可能还不熟，待补）
#include<iostream>
#include<map>
using namespace std;
#define ll long long
const int maxn = 1e5 + 5;
ll a[maxn];
map<ll, int> mp;
int main(){
int n;
cin >> n;
for (int i = 1; i <= n; ++i){
cin >> a[i];
}
int r = n;
while(r >= 1){
if(mp[a[r]] >= 1)
break;
++mp[a[r]];
--r;
}
int ans = INT_MAX;
for (int i = 1; i <= n; ++i){
ans = min(ans, r - i + 1);
++mp[a[i]];
while(mp[a[i]] >= 2 && r <= n ){
++r;
--mp[a[r]];
}
if(mp[a[i]] >= 2)
break;
}
cout << ans;
return 0;
}
CodeForces 993A Two Squares
麻了，暴力枚举每个点。没看到有人过，所以根本没考虑，但凡看下题
用数学下来还错了，麻中麻
还是有点难度，不知道怎么遍历菱形
等我深入学了计算几何吧字母太头疼了
挂几个题解在这里
A
B
C
CodeForces 1B Spreadsheet
推出来是二十六进制转十进制，但是我们三都不会大进制转小进制里面的细节，决定放弃
…题读错了还过了四个样例，我说呢；还错得没法改
对比了下思路，果然又复杂又乱orz
被卡吐了
sscanf需要返回值确定是否转化成功
int main() {
IOS;
int t;
cin >> t;
while(t--){
string s;
cin >> s;
int a, b;
char c[5], d;
if(sscanf(s.c_str(),"R%dC%d", &a, &b) == 2){//这个返回值看对不对
int ans[5], cnt = 0;
while(b != 0){
ans[cnt++] = (b - 1) % 26;
b = (b - 1) / 26;
}
for (int i = cnt - 1; i >= 0; --i){
printf("%c", ans[i] + 'A');
}
printf("%d\n", a);
}
else if(sscanf(s.c_str(),"%[A-Z]%d", &c, &a)){
int ans = 0;
// for (int i = 0; c[i] >= 'A' && c[i] <= 'Z'; ++i){
// ans *= 26;
// ans += (int)(c[i] - 'A' + 1);
// }
char *p;//指针的神奇用法
for(p = c, ans = 0; *p; p++){
ans = ans * 26, ans += *p - 'A' + 1;
}
printf("R%dC%d\n",a, ans);
}
}
return 0;
}
CodeForces 987D Fair
很明显的最短路，和之前写过的一个按权值搜的类似，还剩半小时，回忆+调出来的时间多半不够，放弃了
写倒是容易，调了半天，又发现思路错了…
摆了
下次认真补
int n, m, k, s;
int v[maxn];
vector<int> g[maxn];
vector<int> mp[maxn];
int dist[maxn][110];
bool vis[maxn];
int anss[maxn];
int temps[maxn];
void dij(int st) {
queue<int> q;
re(vis);
remax(temps);
for(int i = 1; i <= n; ++i){
if(v[i] == st) {
q.push(i);
temps[i] = 0;
}
}
while (!q.empty()) {
int id = q.front();
q.pop();
if (vis[id])
continue;
vis[id] = true;
for (auto w : g[id]) {
if (temps[w] > temps[id] + 1) {
temps[w] = temps[id] + 1;
q.push(w);
}
}
}
for(int i = 1; i <= n; ++i){
if(vis[i] != 0x3f3f3f3f){
dist[i][st] = temps[i];
}
}
}
int main() {
IOS;
// freopen("P1908_6.in","r",stdin);//读入数据
// freopen("P1908.out","w",stdout); //输出数据
remax(dist);
cin >> n >> m >> k >> s;
for (int i = 1; i <= n; ++i) {
cin >> v[i];
mp[v[i]].push_back(i);
}
for (int i = 1; i <= m; ++i) {
int a, b;
cin >> a >> b;
g[a].push_back(b);
g[b].push_back(a);
}
for (int i = 1; i <= k; ++i) {
dij(i);
}
for (int i = 1; i <= n; ++i){
sort(dist[i] + 1, dist[i] + k + 1);
int sum = 0;
for (int j = 1; j <= s; ++j){
sum += dist[i][j];
}
cout << sum << " ";
}
return 0;
}
CodeForces 1237B Balanced Tunnel
推了一下，觉得是逆序对，被否了；考虑类似队列+栈的结合，没考虑出来，应该没多难啊
是t宝出的题！
官方题解：记录下每个每个车子出去的时间，如果后面的时间比前面的时间小，说明它超车了。因为先进先出，一定是递减顺序
pair排序加对比取最大值也可以
#include<iostream>
#include<vector>
using namespace std;
const int maxn = 1e5 + 5;
int a[maxn], b[maxn], c[maxn],pos[maxn];
int main(){
int n;
cin >> n;
for(int i = 1; i <= n; ++i){
cin >> a[i];
}
for (int i = 1; i <= n; ++i){
cin >> b[i];
}
for (int i = 1; i <= n; ++i){
pos[b[i]] = i;
}
for (int i = 1; i <= n; ++i){
c[i] = pos[a[i]];
}
int ans = 0;
int t = 0;
for (int i = 1; i <= n; ++i){
if(t > c[i])
++ans;
t = max(t, c[i]);
}
cout << ans;
return 0;
}
总之原理一样，但感觉这个更好理解(?)
#include<iostream>
#include<vector>
#include<algorithm>
using namespace std;
const int maxn = 1e5 + 5;
pair<int, int> a[maxn];
int main(){
int n;
cin >> n;
for(int i = 1; i <= n; ++i){
int t;
cin >> t;
a[t].first = i;
}
for (int i = 1; i <= n; ++i){
int t;
cin >> t;
a[t].second = i;
}
sort(a + 1, a + n + 1);
int ans = 0;
int t = 0;
for (int i = 1; i <= n; ++i){
if(t > a[i].second)
++ans;
t = max(t, a[i].second);
}
cout << ans;
return 0;
}
CodeForces 1499C Minimum Grid Path
感觉像个分配的数学问题，卡在分配了，结果是贪心
一直wa4，是因为INT_MAX还是不够大，换成了0x3f3f3f3f3f3f3f3f过的
结合理解
#include<iostream>
#include<vector>
#include<algorithm>
using namespace std;
const int maxn = 1e5 + 5;
#define ll long long
int main(){
int t;
cin >> t;
while(t--){
int n;
cin >> n;
ll prea[maxn], a[maxn], preb[maxn], prema[maxn], premb[maxn];
for (int i = 1; i <= n; ++i){
cin >> a[i];
if(i & 1){
if(i == 1)
prea[i] = prema[i] = a[i];
else
prema[i] = min(prema[i - 2], a[i]), prea[i] = prea[i - 2] + a[i];
}
else {
if(i == 2)
preb[i] = premb[i] = a[i];
else
premb[i] = min(premb[i - 2], a[i]), preb[i] = preb[i - 2] + a[i];
}
}
ll ans = 0x3f3f3f3f3f3f3f;
for (int i = 2; i <= n; ++i){
ll temp = 0, k = i / 2;
if(i & 1){
temp += prea[i] + (n - k - 1) * prema[i];
temp += preb[i - 1] + (n - k) * premb[i - 1];
}
else {
temp += prea[i - 1] + (n - k) * prema[i - 1];
temp += preb[i] + (n - k) * premb[i];
}
ans = min(ans, temp);
}
cout << ans << endl;
}
return 0;
}
怎么说呢，其他的题看了眼题号，暂不涉及哈
