E1041 计算机学会CSP-J/S -2026年CSP-J模拟卷-2-考试

01 #include <iostream>
02 using namespace std;
03 void hanoi(int n,char a,char b,char c) {
04     if (n == 1)
05         cout << n << " : " << a << " --> " << c << endl;
06     else {
07         hanoi(n-1, a, c, b);
08         cout << n << " : " << a << " --> " << c << endl;
09         hanoi(n-1, b, a, c);
10     }
11 }
12 int main() {
13     int n;
14     cin >> n;
15     hanoi(n, 'A', 'B', 'C');
16     return 0;
17 }

当 n >= 0 时，程序不会出现死循环。

01 #include <cstdio>
02 using namespace std;
03 #define N 1005
04 int num[N];
05 int main() {
06     int a1 = 1, n, x;
07     scanf("%d", &n);
08     num[1] = 1;
09     for (int i=1; i<=n; ++i) {
10         x = 0;
11         for (int j=1; j<=a1; ++j) {
12             num[j] = num[j] * 5 + x;
13             x = num[j] / 10;
14             num[j] %= 10;
15         }
16         if (x > 0) num[++a1] = x;
17     }
18     printf("0");
19     for (int i=a1; i<=n; ++i) putchar('0');
20     for (int i=a1; i>=1; i--) printf("%d", num[i]);
21     putchar('\n');
22     return 0;
23 }

程序输出的是 5^n 的值。

01 #include <iostream>
02 using namespace std;
03 int l,n,m,a [50005], ans;
04 bool check(int dis) {
05     int count=0, last =0;
06     for (int i=1; i<=n+1; i++)
07         if (a[i] - last < dis) count++;
08         else last = a[i];
09     if (count > m) return 0;
10     return 1;
11 }
12 
13 int main() {
14     // 输入保证 l,n,m,a[i] 是正整数，且 a[i] 严格递增
15     cin >> l >> n >> m;
16     for (int i=1; i<=n; i++) cin >> a[i];
17     a[n+1] = l;
18     int fl = 0, fr = l;
19     while (fl <= fr) {
20         int mid = (fl + fr) / 2;
21         if (check(mid)) fl=mid+1, ans=mid;
22         else fr=mid-1;
23     }
24     cout << ans << endl;
25     return 0;
26 }

将 main() 函数中 while 之前的 fl = 0 改为 fl = 1，程序输出结果必定不变。

01 #include <iostream>
02 using namespace std;
03 int main() {
04     int edges;
05     int points;
06     int dis[10];
07     int flag[10];
08     int infinity=9999999;
09     cin >> points >> edges;
10     int edg[10][10];
11     for (int i=1; i<=points; i++) {
12         for (int j=1; j<=points; j++) {
13             if (i == j) {
14                 edg[i][j] = ①;
15             } else {
16                 edg[i][j] = ②;
17             }
18         }
19     }
20     int point1, point2, quanzhi;
21     for (int i=1;i<=edges; i++) {
22         cin >> point1 >> point2 >> quanzhi;
23         edg[point1][point2] = ③;
24     }
25     for (int i=1; i<=points; i++) dis[i] = edg[1][i];
26     for (int i=1; i<=points; i++) flag[i]=0;
27     flag[1]=1;
28     int min,u;
29     for (int i=1; i<=points-1; i++) {
30         // 源点到源点不用比较，因此总的次数少一次
31         min = infinity;
32         for (int j=1; j<=points; j++) {
33             if (flag[j]==0 && dis[j]<min) {
34                 // 核心思想：依次比较出离源点最近的点
35                 min = ④;
36                 u=j;
37             }
38         }
39         flag[u] = 1;
40         for (int v=1; v<=points; v++) {
41             // 找出离源点最近的点后，更新 dis 里面的源点到各个点的值是否最小
42             if (edg[u][v] < infinity) {
43                 if (dis[v] > dis[u] + edg[u][v]) {
44                     dis[v] = ⑤;
45                 }
46             }
47         }
48     }
49     for (int i=1; i<=points; i++) cout << dis[i]<<" ";
50     cout << endl;
51     return 0;
52 }

① 处应填 ( )。

01 #include <iostream>
02 #include <algorithm>
03 using namespace std;
04 int main() {
05     int total_weight=10;
06     int w[6] = {0,5,4,3,2,1};
07     int v[6] = {0,1,2,3,4,5};
08     int dp[11] = { ① };
09     for (int i=1; i<=2; i++)
10         for (int j=w[i]; j<=③; j++)
11             dp[j] = ④;
12     cout << ⑤ << endl;
13     return 0;
14 }

① 处应填 ( )。