1.帮贡排序
解题思路：按照题意，排序模拟即可
#include <iostream>
#include <vector>
#include <algorithm>
#include <string>
using namespace std;
struct Member {
string name;
string position;
int contribution;
int level;
int original_order;
};
int getPriority(const string& pos) {
if (pos == "BangZhu") return 0;
if (pos == "FuBangZhu") return 1;
if (pos == "HuFa") return 2;
if (pos == "ZhangLao") return 3;
if (pos == "TangZhu") return 4;
if (pos == "JingYing") return 5;
return 6; // BangZhong
}
bool compareDisplay(const Member& a, const Member& b) {
int pa = getPriority(a.position);
int pb = getPriority(b.position);
if (pa != pb) return pa < pb;
if (a.level != b.level) return a.level > b.level;
return a.original_order < b.original_order;
}
bool compareContribution(const Member& a, const Member& b) {
if (a.contribution != b.contribution) return a.contribution > b.contribution;
return a.original_order < b.original_order;
}
int main() {
int n;
cin >> n;
vector<Member> members(n);
vector<Member> leaders;
vector<Member> others;
for (int i = 0; i < n; ++i) {
cin >> members[i].name >> members[i].position >> members[i].contribution >> members[i].level;
members[i].original_order = i;
if (members[i].position == "BangZhu" || members[i].position == "FuBangZhu") {
leaders.push_back(members[i]);
} else {
others.push_back(members[i]);
}
}
sort(others.begin(), others.end(), compareContribution);
int huFaCount = 0, zhangLaoCount = 0, tangZhuCount = 0, jingYingCount = 0;
for (auto& member : others) {
if (huFaCount < 2) {
member.position = "HuFa";
huFaCount++;
} else if (zhangLaoCount < 4) {
member.position = "ZhangLao";
zhangLaoCount++;
} else if (tangZhuCount < 7) {
member.position = "TangZhu";
tangZhuCount++;
} else if (jingYingCount < 25) {
member.position = "JingYing";
jingYingCount++;
} else {
member.position = "BangZhong";
}
}
vector<Member> allMembers = leaders;
allMembers.insert(allMembers.end(), others.begin(), others.end());
sort(allMembers.begin(), allMembers.end(), compareDisplay);
for (const auto& member : allMembers) {
cout << member.name << " " << member.position << " " << member.level << endl;
}
return 0;
}2.阶乘数码
解题思路：高精度，单精度问题，套用模版，最后统计一遍即可
#include<bits/stdc++.h>
#define ll long long
#define endl "\n"
using namespace std;
void dis_time() {
ios::sync_with_stdio(0);
cin.tie(0), cout.tie(0);
}
void mult(vector<int>& ans, int x) {
reverse(ans.begin(), ans.end());
int up = 0;
for (int i = 0; i < ans.size(); i++) {
ans[i] = ans[i] * x + up;
up = ans[i] / 10;
ans[i] %= 10;
}
while (up) {
int cur = up % 10;
ans.push_back(cur);
up /= 10;
}
reverse(ans.begin(), ans.end());
}
void solve() {
int n, m;
cin >> n >> m;
vector<int>ans;
ans.push_back(1);
for (int i = 2; i <= n; i++) {
mult(ans, i);
}
int res = 0;
for (auto i : ans) {
if (i == m)
res++;
}
cout << res << endl;
}
int main() {
dis_time();
int t; cin >> t;
while(t--)
solve();
return 0;
}
3.最大乘积
解题思路：对于一个数学结论，如果可以相同经量分成3的多少次方，但是这里不一样，由
可知，尽可能把数变的小，这样就多，乘积也就最大，即用贪心的思路去解题即可
#include <iostream>
#include <vector>
#include <cmath>
#include <cstring>
#include <algorithm>
using namespace std;
const int MAXN = 10003;
const int MAXL = 100003;
const int INF = 2147483647;
struct BigNumber {
int digits[MAXL];
int length;
BigNumber() : length(0) {
memset(digits, 0, sizeof(digits));
}
BigNumber multiply(int num) {
BigNumber result;
for (int i = 1; i <= length; i++) {
result.digits[i] = digits[i] * num;
}
for (int i = 1; i <= length + 10; i++) {
result.digits[i + 1] += result.digits[i] / 10;
result.digits[i] %= 10;
if (result.digits[i]) {
result.length = i;
}
}
return result;
}
};
int main() {
int n;
cin >> n;
double logValues[MAXN];
for (int i = 1; i <= n; i++) {
logValues[i] = log(i);
}
double dp[MAXN];
int path[MAXN];
for (int i = 1; i <= n; i++) {
dp[i] = -INF;
}
dp[0] = 0;
for (int i = 1; i <= n; i++) {
for (int j = n; j >= i; j--) {
if (dp[j - i] + logValues[i] > dp[j]) {
dp[j] = dp[j - i] + logValues[i];
path[j] = j - i;
}
}
}
vector<int> decomposition;
for (int p = n; p != 0; p = path[p]) {
decomposition.push_back(p - path[p]);
}
sort(decomposition.begin(), decomposition.end());
for (int num : decomposition) {
cout << num << " ";
}
cout << endl;
BigNumber product;
product.length = 1;
product.digits[1] = 1;
for (int num : decomposition) {
product = product.multiply(num);
}
for (int i = product.length; i >= 1; i--) {
cout << product.digits[i];
}
cout << endl;
return 0;
}4.[NOIP 1998 提高组] 拼数
解题思路：按照ASCII码来排序，排序一下，输出即可
#include<iostream>
#include<string>
#include<algorithm>
using namespace std;
string s[21];int n;
bool cmp(const string &a,const string &b) {
return (a+b > b+a);
}
int main(void) {
cin >> n;
for(int i=1;i<=n;++i) cin >> s[i];
sort(s+1,s+n+1,cmp);
for (int i=1;i<=n;++i) cout << s[i];
return 0;
}5.统计方形（数据加强版）
解题思路：找规律，对于每个格子，假设都为方形的右下角，对于正方形，有这个格子在内的正方形个数为min（x,y）,对于这个格子的所有的长方形（包括正方形）为x*y,既如此，遍历每个格子为右下角的情况，相加求解
#include<cstdio>
#include<iostream>
#include<cstring>
#include<algorithm>
#define ll long long
using namespace std;
int main(){
ll n,m,i,j,sum=0,sum1=0;
cin>>n>>m;
for(i=1;i<=n;i++){
for(j=1;j<=m;j++){
sum+=min(i,j);
sum1+=i*j;
}
}
cout<<sum<<" "<<sum1-sum<<endl;
return 0;
}6.P1618 三连击（升级版）
解题思路：这个就dfs,找全排列在A:B:C的条件下有多少个
#include<bits/stdc++.h>
#define ll long long
#define endl "\n"
using namespace std;
void dis_time() {
ios::sync_with_stdio(0);
cin.tie(0), cout.tie(0);
}
bool vis[10];
int num[11];
int a, b, c;
int key = 0;
void dfs(int x) {
if (x > 9) {
int num1 = num[1] * 100 + num[2] * 10 + num[3];
int num2 = num[4] * 100 + num[5] * 10 + num[6];
int num3 = num[7] * 100 + num[8] * 10 + num[9];
if (num1 * b == num2 * a && num1 * c == num3 * a) {
key = 1;
cout << num1 << " " << num2 << " " << num3 << endl;
}
return;
}
for (int i = 1; i <= 9; i++) {
if (!vis[i]) {
vis[i] = true;
num[x] = i;
dfs(x + 1);
vis[i] = false;
}
}
}
void solve() {
cin >> a >> b >> c;
dfs(1);
if (!key)
cout << "No!!!" << endl;
}
int main() {
dis_time();
solve();
return 0;
}
6.P3654 First Step (ファーストステップ)
解题思路：对于这个题，可以直接对没列，每行统计就行，但是我最开始写的是dfs，计算出所有的可能然后/2,但是要特判k为1时
#include<bits/stdc++.h>
#define ll long long
#define endl "\n"
using namespace std;
void dis_time() {
ios::sync_with_stdio(0);
cin.tie(0); cout.tie(0);
}
int ans = 0;
char graph[110][110];
int n, m, k;
void dfs(int x, int y) {
if (x >= k) {
int key = 0;
for (int i = x; i > x - k; i--) {
if (graph[i][y] != '.') {
key = 1;
break;
}
}
if (!key)
ans++;
}
if (n - x >= k - 1) {
int key = 0;
for (int i = x; i <= x + k - 1; i++) {
if (graph[i][y] != '.') {
key = 1;
break;
}
}
if (!key)
ans++;
}
if (y >= k) {
int key = 0;
for (int i = y; i > y - k; i--) {
if (graph[x][i] != '.') {
key = 1;
break;
}
}
if (!key)
ans++;
}
if (m - y >= k - 1) {
int key = 0;
for (int i = y; i <= y + k - 1; i++) {
if (graph[x][i] != '.') {
key = 1;
break;
}
}
if (!key)
ans++;
}
}
void solve() {
cin >> n >> m >> k;
for (int i = 1; i <= n; i++) {
for (int j = 1; j <= m; j++) {
cin >> graph[i][j];
}
}
for (int i = 1; i <= n; i++) {
for (int j = 1; j <= m; j++) {
if (graph[i][j] == '.' && k != 1)
dfs(i, j);
}
}
if (k != 1)
cout << ans / 2;
else
cout << ans/4;
}
int main() {
dis_time();
solve();
return 0;
}
6.P1149 [NOIP 2008 提高组] 火柴棒等式
解题思路：就是枚举出每个数字他们所需的火柴个数，然后暴力即可
#include<bits/stdc++.h>
#define ll long long
#define endl "\n"
using namespace std;
void dis_time() {
ios::sync_with_stdio(0);
cin.tie(0); cout.tie(0);
}
int num[2500] = { 6,2,5,5,4,5,6,3,7,6 };
void solve() {
int n; cin >> n;
n -= 4;
for (int i = 10; i <= 2000; i++) {
int temp = i;
while (temp) {
num[i] += num[temp % 10];
temp /= 10;
}
}
int ans = 0;
int t = 0;
for (int i = 0; i <= 1000; i++) {
for (int j = 0; j <= 1000; j++) {
int temp = i + j;
if (num[i] + num[j] + num[temp] == n && i != j) {
ans ++;
}
else if (num[i] + num[j] + num[temp] == n && i == j) {
ans++; t++;
}
}
}
cout << ans - t / 2;
}
int main() {
dis_time();
solve();
return 0;
}
7.P3799 小 Y 拼木棒
解题思路：对于这个题，有a = b = c + d,只要去找a,c就够了,也就是双层for循环，但是常规的双层for循环会时间超限，所有可以用离散化map来存储，元素和元素个数，然后求解
#include <bits/stdc++.h>
#define ll long long
#define endl "\n"
using namespace std;
const int MOD = 1e9 + 7;
void dis_time(){
ios::sync_with_stdio(false);
cin.tie(nullptr);
}
void solve() {
int n;
cin >> n;
unordered_map<int, int> freq;
int max_len = 0;
for (int i = 0; i < n; ++i) {
int temp;
cin >> temp;
freq[temp]++;
if (temp > max_len) {
max_len = temp;
}
}
ll ans = 0;
for (auto& a : freq) {
if (a.second < 2) continue;
ll cnt_a = (ll)a.second * (a.second - 1) / 2 % MOD;
for (auto& c : freq) {
int d = a.first - c.first;
if (d <= 0 || c.first >= d) continue;
if (freq.count(d)) {
ll cnt_cd = (ll)c.second * freq[d] % MOD;
ans = (ans + cnt_a * cnt_cd % MOD) % MOD;
}
}
if (a.first % 2 == 0) {
int c = a.first / 2;
if (freq.count(c) && freq[c] >= 2) {
ll cnt_c = (ll)freq[c] * (freq[c] - 1) / 2 % MOD;
ans = (ans + cnt_a * cnt_c % MOD) % MOD;
}
}
}
cout << ans << endl;
}
int main() {
dis_time();
solve();
return 0;
}
