CCF CSP历年一二题代码汇总

实话说如果不是为了骗访问量，才不会写12题的qwq

 

201803-1 跳一跳

第一题向来是送分题。。。但是注意读题。。。

模拟题按题意走是坠稳的

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<iostream>
#define LL long long
#define debug(x) cout << "[" << x << "]" << endl
using namespace std;

int main(){
    int ans = 0, last = 0, n;
    while (scanf("%d", &n) == 1 && n){
        if (n == 2){
            if (ans == 0 || last == 1){
                last = 2;
                ans += 2;
            }
            else {
                last += 2;
                ans += last;
            }
        }
        else {
            ans++;
            last = 1;
        }
    }
    printf("%d\n", ans);
    return 0;
}

 

201803-2 碰撞的小球

t只有100，n也只有100，tn^2随便模拟一下。。

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<iostream>
#define LL long long
#define debug(x) cout << "[" << x << "]" << endl
using namespace std;

int a[110], b[110];

int main(){
    int n, l, t;
    scanf("%d%d%d", &n, &l, &t);
    for (int i = 1; i <= n; i++){
        scanf("%d", &a[i]);
        b[i] = 1;
    }
    while (t--){
        for (int i = 1; i <= n; i++){
            for (int j = i+1; j <= n; j++){
                if (i == j) continue;
                if (a[i] == a[j]){
                    b[i] = -b[i];
                    b[j] = -b[j];
                }
            }
            if (a[i] == 0 || a[i] == l) b[i] = -b[i];
        }
        for (int i = 1; i <= n; i++) a[i] += b[i];
    }
    for (int i = 1; i <= n; i++)
        printf("%d%c", a[i], i == n ? '\n' : ' ');
    return 0;
}

 

201712-1 最小差值

暴力或排序随便怎么实现都行

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<iostream>
#define INF 0x3f3f3f3f
#define LL long long
#define debug(x) cout << "[" << x << "]" << endl
using namespace std;

int a[1010];

int main(){
    int n;
    scanf("%d", &n);
    for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
    sort(a+1, a+n+1);
    int ans = INF;
    for (int i = 1; i < n; i++)
        ans = min(ans, a[i+1]-a[i]);
    printf("%d\n", ans);
    return 0;
}

 

201712-2 游戏

写个死循环模拟到只剩一个人为止

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<iostream>
#define LL long long
#define debug(x) cout << "[" << x << "]" << endl
using namespace std;

bool vis[1010];

int main(){
    int n, k, num = 0, sum = 0;
    scanf("%d%d", &n, &k);
    while (1){
        for (int i = 1; i <= n; i++){
            if (vis[i]) continue;
            num++;
            if (num%10 == k || num%k == 0) {
                vis[i] = 1;
                sum++;
            }
            if (sum == n-1) break;
        }
        if (sum == n-1) break;
    }
    for (int i = 1; i <= n; i++){
        if (!vis[i]){
            printf("%d\n", i);
            break;
        }
    }
    return 0;
}

 

201709-1 打酱油

算就行

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<iostream>
#define LL long long
#define debug(x) cout << "[" << x << "]" << endl
using namespace std;

int main(){
    int n, ans = 0;
    scanf("%d", &n);
    n /= 10;
    ans += n/5*7;
    n %= 5;
    ans += n/3*4;
    n %= 3;
    ans += n;
    printf("%d\n", ans);
    return 0;
}

 

201709-2 公共钥匙盒

看题第一反应线段树做RMQ+树上的二分，看数据范围沉默。。。

此处只需要优先队列存时间点然后暴力模拟，注意重载的优先级

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<iostream>
#include<queue>
#include<cstdlib>
#define LL long long
#define debug(x) cout << "[" << x << "]" << endl
using namespace std;

struct node{
    int id, s, op;
    bool operator < (const node& a) const {
        return s > a.s || ((s == a.s && op < a.op) || (s == a.s && op == a.op && id > a.id));
    }
    node(int id = 0, int s = 0, int op = 0): id(id), s(s), op(op){}
};
int a[10010];

int main(){
    priority_queue<node> q;
    int n, k, w, s, c;
    scanf("%d%d", &n, &k);
    for (int i = 1; i <= k; i++){
        scanf("%d%d%d", &w, &s, &c);
        q.push(node(w, s, 0));
        q.push(node(w, s+c, 1));
    }
    for (int i = 1; i <= n; i++) a[i] = i;
    while (!q.empty()){
        node u = q.top(); q.pop();
        if (u.op == 1){
            for (int i = 1; i <= n; i++){
                if (a[i] == 0){
                    a[i] = u.id;
                    break;
                }
            }
        }
        else {
            for (int i = 1; i <= n; i++){
                if (u.id == a[i]){
                    a[i] = 0;
                    break;
                }
            }
        }
    }
    for (int i = 1; i <= n; i++)
        printf("%d%c", a[i], i == n ? '\n' : ' ');
    return 0;
}