标签：PAT 2021.9 scanf ++ int vector ans include 刷题

1001 考察点 字符串的处理

#include<iostream>
#include<vector>
#include<cmath>
#include<string>
using namespace std;

int main(){
    int x, y;
    cin >> x >> y;
    string s = to_string(x+y);
    for(int i = 0; i < s.size(); i++){
        printf("%c", s[i]);
        if(s[i]=='-') continue;
        int t = s.size()-i-1;
        if(t%3==0 && t!=0) printf(",");
    }
    printf("\n");
}

1002 思路：开个一百的数组，记录每一项的系数。最后数下几个非零项，依次输出

#include<iostream>
#include<vector>
using namespace std;
const int M = 1000+3;

int main(){
    vector<double> v(M, 0);
    int K; 
    scanf("%d", &K);
    for(int i = 0; i < K; i++){
        int x; double y;
        scanf("%d %lf", &x, &y);
        v[x] += y;
    }
    int cnt = 0;
    scanf("%d", &K);
    for(int i = 0; i < K; i++){
        int x; double y;
        scanf("%d %lf", &x, &y);
        v[x] += y;
    }
    for(int i = M-1; i >= 0; i--)
        if(v[i]!=0) cnt++;
    printf("%d", cnt);
    for(int i = M-1; i >= 0; i--){
        if(v[i] !=0 ) printf(" %d %.1f", i, v[i]);
    }
    printf("\n");
}

 1003 Dij那什么求最短路的算法。要新建两个变量保存路径条数和最大重量，更新最短路的时候分为大于和等于两种情况。

#include<iostream>
#include<vector>
#include<limits.h>
using namespace std;
int n, m, c1, c2;
vector<int> w;
vector<vector<int>> dis;
vector<int> visited;
vector<int> cnts;
vector<int> weights;
int main(){
    scanf("%d %d %d %d", &n, &m, &c1, &c2);
    for(int i = 0; i < n; i++){
        int x; scanf("%d", &x);
        w.push_back(x);
    }
    dis.resize(n, vector<int>(n, -1));
    for(int i = 0; i < m; i++){
        int x, y, d; scanf("%d %d %d", &x, &y, &d);
        dis[x][y] = dis[y][x] = d;
    }
    dis[c1][c1] = 0;
    visited.resize(n, 0);
    //visited[c1] = 1;
    cnts.resize(n, 0); cnts[c1] = 1;
    weights.resize(n, 0); weights[c1] = w[c1];
    for(int i = 0; i < n; i++){
        int M = INT_MAX;
        int t = -1;
        for(int j = 0; j < n; j++){
            if(visited[j]==0 && dis[c1][j]!=-1 && dis[c1][j]<M){
                M = dis[c1][j];
                t = j;
            }
        }
        visited[t] = 1;
        for(int j = 0; j < n; j++){
            if(visited[j]==0 && dis[t][j]!=-1){
                if(dis[c1][j]==-1 || dis[c1][j]>dis[c1][t]+dis[t][j]){
                    dis[c1][j] = dis[c1][t]+dis[t][j];
                    cnts[j] = cnts[t];
                    weights[j] = weights[t]+w[j];
                }
                else if(dis[c1][j]==dis[c1][t]+dis[t][j]){
                    cnts[j] += cnts[t];
                    weights[j] = max(weights[j], weights[t]+w[j]);
                }
            }
        }
    }
    printf("%d %d\n", cnts[c2], weights[c2]);
    //cout << cnts[c2] << " ;
}

1004 bfs

#include <iostream>
#include <vector>
#include<queue>
#include<unordered_map>
using namespace std;
const int MAX = 100+2;
int N, M;

int main(){
    scanf("%d %d", &N, &M);
    vector<vector<int>> tree(MAX);
    for(int i = 0; i < M; i++){
        int id, k;
        scanf("%d %d", &id, &k);
        for(int j = 0; j < k; j++){
            int x; scanf("%d", &x);
            tree[id].push_back(x);
        }
    }
    int root = 1;
    queue<int> q;
    vector<int> cnt(MAX, 0);
    unordered_map<int, int> deep; 
    deep[1] = 0; q.push(1);
    int D = 0;
    while(!q.empty()){
        int t = q.front(); q.pop();
        D = deep[t];
        if(tree[t].empty()) cnt[deep[t]]++;
        for(int i = 0; i < tree[t].size(); i++){
            int y = tree[t][i];
            deep[y] = deep[t]+1;
            q.push(y);
        }
    }
    bool first = true;
    for(int i = 0; i <= D; i++)
    {
        if(first) first = false; else printf(" ");
        printf("%d", cnt[i]);
    }
    printf("\n");
    
    
}

 1005 注意和直接为零的情况。WA了一个测试点。

#include<iostream>
#include<vector>
using namespace std;
vector<string> v = {"zero", "one", "two", "three", "four", "five", "six", "seven", "eight", "nine"};
int main(){
    string s;
    cin >> s;
    int n = 0;
    for(int i = 0; i < s.size(); i++){
        n += s[i]-'0';
    }
    if(n==0) {
        cout << "zero" << endl;
        return 0;
    }
    bool first = true;
    vector<string> ans;
    while(n>0){
        //
        ans.push_back(v[n%10]);
        n = n/10;
    }
    for(int i = ans.size()-1; i>=0; i--){
        if(first) first = false; else printf(" ");
        cout << ans[i];
    }
    
    printf("\n");
}

 1006 就按题目说的做

#include<iostream>
#include<vector>
using namespace std;
int m;

int main(){
    scanf("%d", &m);
    vector<string> ids;
    //vector<int> arrs, leas;
    int A = 1e8, B = 0;
    int x = -1, y = -1;
    for(int i = 0; i < m; i++){
        string s; cin >> s;
        ids.push_back(s);
        int x1, y1, z1, x2, y2, z2;
        scanf("%d:%d:%d %d:%d:%d", &x1, &y1, &z1, &x2, &y2, &z2);
        //arrs.push_back(x1*3600+y1*60+z1);
        //leas.push_back(x2*3600+y2*60+z2);
        int a = x1*3600+y1*60+z1, b = x2*3600+y2*60+z2;
        if(a<A) {A = a; x = i;}
        if(b>B) {B = b; y = i;}
    }
    cout << ids[x] << " " << ids[y] << endl;
    
}

 1007 滑动窗口 保持窗口内和大于零，看情况更新答案。注意全负数的情况，还要注意和为零算普通的答案。

#include<iostream>
#include<vector>
using namespace std;
int m;
vector<int> v;
int main(){
    scanf("%d", &m);
    for(int i = 0; i < m; i++){
        int x; scanf("%d", &x);
        v.push_back(x);
    }
    int l = 0, r = 0;
    int sum = 0;
    int L = -1, R = -1, ans = -1;
    while(r < m){
        sum += v[r];
        if(sum < 0) {
            l = r+1;
            sum = 0;
        }
        else if(sum > ans){
            ans = sum;
            L = l; R = r;
            //cout << l << " " << r << " " << sum << endl;
        }
        r++;
    }
    if(ans == -1) printf("%d %d %d\n", 0, v[0], v[m-1]);
    else printf("%d %d %d\n", ans, v[L], v[R]);
}

1008 题面意思

#include<iostream>
#include<vector>
using namespace std;
int n;
vector<int> v;
int main(){
    scanf("%d", &n);
    for(int i = 0; i < n; i++){
        int x; scanf("%d", &x);
        v.push_back(x);
    }
    int ans = 0;
    for(int i = 0; i < n; i++){
        if(i==0) ans += v[i]*6;
        else {
            if(v[i] > v[i-1]) 
                ans += (v[i]-v[i-1])*6;
            else 
                ans += (v[i-1]-v[i])*4;
        }
        ans += 5;
    }
    printf("%d\n", ans);
    
}

1009 和上面那个多项式题目类似。哇照这个进度我一天能不能A十道啊。

#include<iostream>
#include<vector>
#include<unordered_map>
using namespace std;
const int M = 2000+5;
int k;

int main(){
    unordered_map<int, double> T1, T2;
    scanf("%d", &k);
    for(int i = 0; i < k; i++){
        int x; double y; 
        scanf("%d %lf", &x, &y);
        T1[x] = y;
    }
    scanf("%d", &k);
    for(int i = 0; i < k; i++){
        int x; double y;
        scanf("%d %lf", &x, &y);
        T2[x] = y;
    }
    vector<double> cos(M, 0);
    for(const auto& a:T1){
        for(const auto& b:T2){
            cos[a.first+b.first] += a.second*b.second;
        }
    }
    int cnt = 0;
    for(int i = 0; i < M; i++)
        if(cos[i]!=0) cnt++;
    printf("%d", cnt);
    for(int i = M-1; i >= 0; i--){
        if(cos[i]!=0) printf(" %d %.1f", i, cos[i]);
    }
    printf("\n");
}

1010 救命 WA了两次呜呜呜还没写出来待我看看。错误点1：进制选择没有上限，我自以为地设置为36。看了一圈题解要二分还要处理溢出orz明天再弄吧。但是为了一天A10题我决定苟个下一题。

1011

#include<iostream>
#include<vector>
using namespace std;

int main(){
    double x;
    double ans = 1;
    vector<char> v = {'W', 'T', 'L'};
    for(int i = 0; i < 3; i++){
        x = 0; char t;
        for(int j = 0; j < 3; j++){
            double c; scanf("%lf", &c);
            if(c > x) t = v[j];
            x = max(x, c);
        }
        printf("%c ", t);
        ans *= x;
    }
    ans = (ans*0.65-1)*2;
    printf("%.2f\n", ans);
}

标签：PAT,2021.9,scanf,++,int,vector,ans,include,刷题
来源： https://blog.csdn.net/qq_41922663/article/details/119087516

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