7-1谁先倒
题意：本人语文功底太差，有些许不是很明白，但是这个是这个水题，也没必要深究
#include <iostream>
using namespace std;
int A,B;
int n;
int a,b,c,d;
int j,y;
int main()
{
cin >> A >> B;
cin >> n;
int ying = 0;
while(n --)
{
cin >> a >> b >> c >> d;
int e = a + c;
if(ying != 0){
continue;
}
if(e == b && e == d){
j++;
y++;
}
else if(e == b){
j++;
}
else if(e == d){
y++;
}
if(j >= A){
ying = 2;
}
if(j >= B){
ying = 1;
}
}
if(ying == 1){
cout << 'A' << endl;
cout << y << endl;
}
else{
cout << 'B' << endl;
cout << j << endl;
}
return 0;
}
结果只是得了14分，主要是题意还是没读明白
7-2 重要的话说三遍
题意：把一串字符串输出三遍就好了
#include <iostream>
using namespace std;
int A,B;
int n;
int a,b,c,d;
int j,y;
int main()
{
printf("I'm gonna WIN!\n");
printf("I'm gonna WIN!\n");
printf("I'm gonna WIN!\n");
return 0;
}
7-3 奇偶分家
题意：给定N个正整数，请统计奇数和偶数各有多少个？
#include <iostream>
using namespace std;
int main()
{
int n;
int odd = 0;
cin >> n;
for(int i = 0;i < n;i ++)
{
int x;
cin >> x;
if(x & 1){
odd++;
}
}
cout << odd << ' ' << n - odd << endl;
return 0;
}
7- 4 输出GPLT
题意：给定一个长度不超过10000的、仅由英文字母构成的字符串。请将字符重新调整顺序，按GPLTGPLT…这样的顺序输出，并忽略其它字符。当然，四种字符（不区分大小写）的个数不一定是一样多的，若某种字符已经输出完，则余下的字符仍按GPLT的顺序打印，直到所有字符都被输出。
/*
统计出gplt每个字符的个数，然后按照顺序有啥输出啥就OK了
*/
#include <iostream>
using namespace std;
int a,b,c,d;
int main()
{
string str;
cin >> str;
for(int i = 0;str[i];i ++)
{
if(str[i] == 'G' || str[i] == 'g'){
a++;
}
else if(str[i] == 'P' || str[i] == 'p'){
b++;
}
else if(str[i] == 'L' || str[i] == 'l'){
c++;
}
else if(str[i] == 'T' || str[i] == 't'){
d++;
}
}
while(1)
{
if(a){
a--;
cout << 'G';
}
if(b){
b--;
cout << 'P';
}
if(c){
c--;
cout << 'L';
}
if(d){
d--;
cout << 'T';
}
if(a == 0 && b == 0 && c == 0 && d == 0){
break;
}
}
return 0;
}
7- 5 点赞
题意：点赞
#include <iostream>
#include <unordered_map>
using namespace std;
int a[10000];
int main()
{
int n;
cin >> n;
int m = 0,id;
while(n --)
{
int x,y;
cin >> y;
while(y --)
{
cin >> x;
a[x]++;
if(a[x] > m){
m = a[x];
id = x;
}
if(a[x] == m){
if(x > id){
id = x;
}
}
}
}
cout << id << ' ' << m << endl;
return 0;
}
7-6 情人节 (15分)
题意： 情人节
#include <iostream>
#include <vector>
using namespace std;
vector<string>v;
int main()
{
string str;
while(cin >> str && str != ".")
{
v.push_back(str);
}
if(v.size() >= 14){
cout << v[1] << " and " << v[13] << " are inviting you to dinner..." << endl;
}
else if(v.size() < 2){
cout << "Momo... No one is for you ..." << endl;
}
else{
cout << v[1] << " is the only one for you..." << endl;
}
return 0;
}
7-7 最佳情侣身高差
题意：7-7 最佳情侣身高差
#include <iostream>
#include <vector>
using namespace std;
int main()
{
int n;
char sex;
double he;
cin >> n;
while(n --)
{
cin >> sex >> he;
if(sex == 'M'){
he = he / 1.09;
}
else{
he = he * 1.09;
}
printf("%.2lf\n",he);
}
return 0;
}
7-8 宇宙无敌大招呼 (5分)
题意：7-8 宇宙无敌大招呼
#include <iostream>
#include <vector>
using namespace std;
int main()
{
string name;
cin >> name;
cout << "Hello " << name << endl;
return 0;
}
7-9 排座位 (25分)
题意：如果两位宾客之间是朋友，且没有敌对关系，则输出No problem；如果他们之间并不是朋友，但也不敌对，则输出OK；如果他们之间有敌对，然而也有共同的朋友，则输出OK but…；如果他们之间只有敌对关系，则输出No way。
/*
用一个二维数组来存储敌对关系，用并查集来存储朋友之间的关系，如果两个人有共同好友，那么两个人肯定在一个集合中
*/
#include <iostream>
#include <vector>
using namespace std;
const int N = 110;
int p[N];
int g[N][N];
int find(int x)
{
if(p[x] != x){
p[x] = find(p[x]);
}
return p[x];
}
int n,m,k;
int main()
{
cin >> n >> m >> k;
for(int i = 1;i < N;i ++)
{
p[i] = i;
}
for(int i = 0;i < m;i ++)
{
int a,b,c;
cin >> a >> b >> c;
int fa = find(a);
int fb = find(b);
g[a][b] = g[b][a] = c;
if(c == 1){
p[fa] = fb;
}
}
for(int i = 0;i < k;i ++)
{
int a,b;
cin >> a >> b;
int fa = find(a);
int fb = find(b);
if(fa == fb && g[a][b] != -1) {
cout << "No problem" << endl;
}
else if(fa != fb && g[a][b] != -1){
cout << "OK" << endl;
}
else if(fa == fb && g[a][b] == -1){
cout << "OK but..." << endl;
}
else if(fa != fb && g[a][b] == -1){
cout << "No way" << endl;
}
}
return 0;
}
7-10 最长对称子串 (25分)
题意：对给定的字符串，本题要求你输出最长对称子串的长度。例如，给定Is PAT&TAP symmetric?，最长对称子串为s PAT&TAP s，于是你应该输出11。
#include <iostream>
#include <cstring>
#include <algorithm>
using namespace std;
string str;
bool check(int l,int r)
{
while(l < r)
{
if(str[l] != str[r]){
return false;
}
l ++;
r --;
}
return true;
}
int main()
{
int maxn = 1;
getline(cin,str);
int len = str.size();
if(str.size() == 0){
cout << 0 << endl;
return 0;
}
for(int i = 0;i < len;i ++)
{
for(int j = len - 1;j > i;j --)
{
if(j - i + 1 < maxn){
break;
}
if(str[j] == str[i]){
if(check(i,j)){
maxn = j - i + 1;
}
}
}
}
cout << maxn << endl;
return 0;
}
7-12 分而治之 (25分)
题意：简单
/*
简单dfs
*/
#include <iostream>
#include <cstring>
#include <algorithm>
#include <unordered_set>
#include <vector>
using namespace std;
const int N = 1e4 + 10;
int n,m,k;
vector<int>v[N];
unordered_set<int>st;
bool dfs(int u)
{
for(auto &x : v[u])
{
if(!st.count(x)){
return true;
}
}
return false;
}
bool check()
{
for(int i = 1;i <= n;i ++)
{
if(st.count(i)){
continue;
}
if(dfs(i)){
return false;
}
}
return true;
}
int main()
{
cin >> n >> m;
for(int i = 0;i < m;i ++)
{
int a,b;
cin >> a >> b;
v[a].push_back(b);
v[b].push_back(a);
}
cin >> k;
while(k --)
{
st.clear();
int t;
cin >> t;
while(t --)
{
int x;
cin >> x;
st.insert(x);
}
if(check()){
cout << "YES" << endl;
}
else{
cout << "NO" << endl;
}
}
return 0;
}
7-13 垃圾箱分布
题意：lj输入，恶心心，就是dijkstra，当然我喜欢spfa来求最短路
#include <iostream>
#include <cstring>
#include <algorithm>
#include <sstream>
#include <vector>
#include <queue>
using namespace std;
const int N = 1020;
typedef pair<int,int> PII;
vector<PII>v[N];
int n,m,k,d;
int dist[N];
bool st[N];
void spfa(int u)
{
memset(dist,0x3f,sizeof dist);
dist[u] = 0;
queue<int>q;
q.push(u);
while(q.size())
{
auto t = q.front();q.pop();
st[t] = false;
for(auto &x : v[t])
{
if(dist[x.first] > dist[t] + x.second){
dist[x.first] = dist[t] + x.second;
if(!st[x.first]){
st[x.first] = true;
q.push(x.first);
}
}
}
}
}
int main()
{
cin >> n >> m >> k >> d;
for(int i = 0;i < k;i ++)
{
string str1,str2;int c;
cin >> str1 >> str2 >> c;
if(str1[0] == 'G' && str2[0] == 'G'){
int s1;
if(str1.size() == 3){
s1 = n + 10;
}
else{
s1 = n + str1[1] - '0';
}
int s2;
if(str2.size() == 3){
s2 = n + 10;
}
else{
s2 = n + str2[1] - '0';
}
v[s1].push_back({s2,c});
v[s2].push_back({s1,c});
}
else if(str1[0] == 'G'){
stringstream ss(str2);
int s2;
ss >> s2;
int s1;
if(str1.size() == 3){
s1 = n + 10;
}
else{
s1 = n + str1[1] - '0';
}
v[s1].push_back({s2,c});
v[s2].push_back({s1,c});
}
else if(str2[0] == 'G'){
stringstream ss(str1);
int s1;
ss >> s1;
int s2;
if(str2.size() == 3){
s2 = n + 10;
}
else{
s2 = n + str2[1] - '0';
}
v[s1].push_back({s2,c});
v[s2].push_back({s1,c});
}
else{
stringstream ss1(str1);
stringstream ss2(str2);
int s1,s2;
ss1 >> s1;
ss2 >> s2;
v[s1].push_back({s2,c});
v[s2].push_back({s1,c});
}
}
int ans = 0;
double ave = 0;
int id = 0;
for(int i = 1;i <= m;i ++)
{
spfa(n + i);
int maxn = 0;
int minn = 1e9;
int sum = 0;
for(int j = 1;j <= n;j ++)
{
maxn = max(maxn,dist[j]);
minn = min(minn,dist[j]);
sum += dist[j];
// cout << dist[j] << ' ';
}
// cout << sum << endl;
if(maxn > d){
continue;
}
else{
if(minn > ans){
ans = minn;
ave = sum;
id = i;
}
else if(minn == ans){
if(ave > sum){
ans = minn;
ave = sum;
id = i;
}
}
}
}
if(id == 0){
cout << "No Solution" << endl;
}
else{
cout << 'G' << id << endl;
double avg = ave * 1.0 / (n);
printf("%.1lf %.1lf\n",ans * 1.0,avg);
}
return 0;
}
7-14 社交集群
题意：晦涩难懂,就是你想的那种暴力，纯暴力，纯暴力，然后就过了，正解我觉得应该是并查集
#include <iostream>
#include <cstring>
#include <algorithm>
#include <set>
#include <vector>
using namespace std;
const int N = 1100;
int n,x,y;
vector<int>v[N];
set<int>stt;
vector<int>res;
bool st1[N];
int main()
{
cin >> n;
for(int i = 1;i <= n;i ++)
{
scanf("%d:",&x);
for(int j = 0;j < x;j ++)
{
cin >> y;
v[i].push_back(y);
}
stt.insert(i);
}
while(1)
{
set<int>st;
int cnt = 1;
for(int i = 1;i <= n;i ++)
{
if(st1[i] == false){
for(auto &x : v[i])
{
st.insert(x);
}
stt.erase(i);
st1[i] = true;
break;
}
}
if(st.size() == 0){
break;
}
for(int i = 1;i <= n;i ++)
{
if(st1[i]){
continue;
}
for(auto &x : v[i]){
if(st.count(x)){
cnt++;
for(auto &x : v[i]){
st.insert(x);
}
stt.erase(i);
st1[i] = true;
for(auto &x : stt)
{
i = x - 1;
break;
}
break;
}
}
}
res.push_back(cnt);
}
cout << res.size() << endl;
sort(res.begin(),res.end());
reverse(res.begin(),res.end());
bool tf = false;
for(auto &x : res)
{
if(tf){
cout << ' ';
}
tf = true;
cout << x;
}
return 0;
}
