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Solution to the fourth "intelligence Cup" National College Students' IT skills competition (group B of the final)
2022-04-23 15:57:00 【Ten thousand volt little sun】
T229470 A. Xiao Zhi's doubts
call string Of substr Method , Violence is enough .
#include<bits/stdc++.h>
#define int long long
#define rep(i, l, r) for (int i = l; i <= r; ++i)
using namespace std;
typedef long long ll;
const int N = 2e6 + 7;
const int mod = 1e9 + 7;
signed main(){
string s;
cin>>s;
int n=s.size();
int ans=0;
for(int i=0;i<n-7;i++){
// if(s[i]=='c'&&s[i+1]=='h'&&s[i+2]=='u'&&s[i+3]&&s[i+4])
string t=s.substr(i,8);
if(t=="chuanzhi") ans++;
}
cout<<ans<<endl;
return 0;
}
T229471 B. A triple
Triple cycle violence
#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const int N = 2e6 + 7;
const int mod = 1e9 + 7;
const int MOD = 998244353;
#define int long long
#define rep(i, l, r) for (int i = l; i <= r; ++i)
int a[N];
void solve(){
int n;
cin>>n;
for(int i=0;i<n;i++){
cin>>a[i];
}
int ans=0;
for(int i=0;i<n;i++){
for(int j=i;j<n;j++){
for(int k=j;k<n;k++){
if(a[i]+a[j]==a[k]) ans++;
}
}
}
cout<<ans<<endl;
}
signed main(){
int _;
cin>>_;
while(_--) solve();
return 0;
}
T229472 C. Line up
As long as another array has the number of that array , In exchange for n 2 n^2 n2 Once, you can , Line up as you want , Bubbling .
But I misunderstood this question , I changed b Array . The title means to change a Array ,wa3
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const int N = 2e6 + 7;
const int mod = 1e9 + 7;
const int MOD = 998244353;
#define int long long
#define rep(i, l, r) for (int i = l; i <= r; ++i)
int a[N], b[N], c[N], d[N];
void solve() {
int n;
cin >> n;
for (int i = 0; i < n; i++) {
cin >> a[i];
c[i]=a[i];
}
for (int j = 0; j < n; j++) {
cin >> b[j];
d[j]=b[j];
}
sort(c,c+n);
sort(d,d+n);
for(int i=0;i<n;i++){
if(c[i]!=d[i]){
cout<<"NO"<<endl;
return;
}
}
vector<int> ans;
vector<int> res;
for (int i = 0; i < n; i++) {
if (a[i] == b[i])
continue;
int pos = -1;
for (int j = i + 1; j < n; j++) {
if (a[j] == b[i]) {
pos = j;
}
}
for (int j = pos; j > i; j--) {
ans.push_back(j+1);
res.push_back(j);
swap(a[j], a[j - 1]);
}
}
cout << "YES" << endl;
for (int i = 0; i < ans.size(); i++) {
cout << ans[i] << " " << res[i] << endl;
}
cout << "0 0\n";
}
signed main() {
int _;
cin >> _;
while (_--)
solve();
return 0;
}
R71214110 Memorizing words
Two points answer m,check Words , Let's start with the first few , Judge how many days he will recite , and k Comparison .
leetcode The original title is ,leetcode410https://leetcode-cn.com/problems/split-array-largest-sum/
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const int N = 2e6 + 7;
const int mod = 1e9 + 7;
const int MOD = 998244353;
#define int long long
#define rep(i, l, r) for (int i = l; i <= r; ++i)
int a[N];
bool check(vector<int> &nums, int x, int m) {
long long sum = 0;
int cnt = 1;
for (int i = 0; i < nums.size(); i++) {
if (sum + nums[i] > x) {
cnt++;
sum = nums[i];
} else {
sum += nums[i];
}
}
return cnt <= m;
}
int work(vector<int> &nums, int m) {
long long left = 0, right = 0;
for (int i = 0; i < nums.size(); i++) {
right += nums[i];
if (left < nums[i]) {
left = nums[i];
}
}
while (left < right) {
long long mid = (left + right) >> 1;
if (check(nums, mid, m)) {
right = mid;
} else {
left = mid + 1;
}
}
return left;
}
void solve() {
int n, k;
cin >> n >> k;
vector<int> nums(n);
for (int i = 0; i < n; i++) {
cin >> nums[i];
nums[i] *= nums[i];
}
cout<<work(nums,k)<<endl;
}
signed main() {
solve();
return 0;
}
T229475 F1. Living in a tree (easy version)
According to the nature of the xor ,a To b XOR of a To c Exclusive or of . This middle point can be taken at will , So let's take a root node . If there is xor[u] ^ xor[v]== k.
Treelike DFS.
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const int N = 5e5 + 7;
const int mod = 1e9 + 7;
const int MOD = 998244353;
#define int unsigned long long
#define rep(i, l, r) for (int i = l; i <= r; ++i)
vector<pair<int, int> > g[N];
int xorr[N];
void dfs(int u, int f, int val) {
xorr[u] = val;
for (int i = 0; i < g[u].size(); i++) {
int next = g[u][i].first, w = g[u][i].second;
if (next == f)
continue;
dfs(next, u, w ^ val);
}
}
void solve() {
int n, m, u, v, w, k;
cin >> n >> m;
for (int i = 1; i < n; i++) {
cin >> u >> v >> w;
g[u].push_back({
v, w});
g[v].push_back({
u, w});
}
dfs(n-1, 0, 0);
while (m--) {
cin >> u >> v >> k;
if ((xorr[u] ^ xorr[v]) == k) {
cout << "YES\n";
} else {
cout << "NO\n";
}
}
}
signed main() {
ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);
solve();
return 0;
}
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本文为[Ten thousand volt little sun]所创,转载请带上原文链接,感谢
https://yzsam.com/2022/04/202204231405457118.html
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