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以a为起点,对右边的无向图进行深度优先遍历,则b、c、d、e四个点中有可能作为最后一个遍历到的点的个数为( )。
![CSP-J1入门级初赛试卷[2021] CSP-J1入门级初赛试卷[2021]](/oj/ueditor/php/upload/image/20220427/1651023937505181.png)
有四个人要从A点坐一条船过河到B点,船开始在A点,该船一次最多可坐两个人。己知这四个人中每个人独自坐船的过河时间分别为1,2,4,8,且两个人坐船的过河时间为两人独自过河时间的较大者。则最短( )时间可以让四个人都过河到B点(包括从B点把船开回A点的时间)。
#include <stdio.h>
int n;
int a[1000];
int f(int x)
{
int ret = 0;
for (; x; x &= x - 1) ret++;
return ret;
}
int g(int x)
{
return x & -x;
}
int main()
{
scanf("%d",&n);
for (int i = 0; i < n; i++) scanf("%d",&a[i]);
for (int i = 0; i < n; i++)
printf("%d",f(a[i])+g(a[i]));
printf("\n");
return 0;
}输入的n等于1001时,程序不会发生下标越界。
#include <stdio.h>
int n;
int a[1000];
int f(int x)
{
int ret = 0;
for (; x; x &= x - 1) ret++;
return ret;
}
int g(int x)
{
return x & -x;
}
int main()
{
scanf("%d",&n);
for (int i = 0; i < n; i++) scanf("%d",&a[i]);
for (int i = 0; i < n; i++)
printf("%d",f(a[i])+g(a[i]));
printf("\n");
return 0;
}输入的a[i]必须全为正整数,否则程序将陷入死循环。
#include <stdio.h>
int n;
int a[1000];
int f(int x)
{
int ret = 0;
for (; x; x &= x - 1) ret++;
return ret;
}
int g(int x)
{
return x & -x;
}
int main()
{
scanf("%d",&n);
for (int i = 0; i < n; i++) scanf("%d",&a[i]);
for (int i = 0; i < n; i++)
printf("%d",f(a[i])+g(a[i]));
printf("\n");
return 0;
}当输入“5 2 11 9 16 10”时,输出为“3 4 3 17 5”。
#include <stdio.h>
int n;
int a[1000];
int f(int x)
{
int ret = 0;
for (; x; x &= x - 1) ret++;
return ret;
}
int g(int x)
{
return x & -x;
}
int main()
{
scanf("%d",&n);
for (int i = 0; i < n; i++) scanf("%d",&a[i]);
for (int i = 0; i < n; i++)
printf("%d",f(a[i])+g(a[i]));
printf("\n");
return 0;
}当输入为“1 511998”时,输出为“18”。
#include <stdio.h>
int n;
int a[1000];
int f(int x)
{
int ret = 0;
for (; x; x &= x - 1) ret++;
return ret;
}
int g(int x)
{
return x & -x;
}
int main()
{
scanf("%d",&n);
for (int i = 0; i < n; i++) scanf("%d",&a[i]);
for (int i = 0; i < n; i++)
printf("%d",f(a[i])+g(a[i]));
printf("\n");
return 0;
}将源代码中g函数的定义(13-16行)移到main函数的后面,程序可以正常编译运行。
#include <stdio.h>
int n;
int a[1000];
int f(int x)
{
int ret = 0;
for (; x; x &= x - 1) ret++;
return ret;
}
int g(int x)
{
return x & -x;
}
int main()
{
scanf("%d",&n);
for (int i = 0; i < n; i++) scanf("%d",&a[i]);
for (int i = 0; i < n; i++)
printf("%d",f(a[i])+g(a[i]));
printf("\n");
return 0;
}当输入为“2 -65536 2147483647”时,输出为( )。
#include <stdio.h>
#include <string.h>
char base[64];
char table[256];
char str[256];
char ans[256];
void init()
{
for (int i = 0; i < 26; i++) base[i] = 'A' + i;
for (int i = 0; i < 26; i++) base[26 + i] = 'a' + i;
for (int i = 0; i < 10; i++) base[52 + i] = '0' + i;
base[62] = '+', base[63] = '/';
for (int i = 0; i < 256; i++) table[i] = 0xff;
for (int i = 0; i < 64; i++) table[base[i]] = i;
table['='] = 0;
}
void decode(char *str)
{
char *ret = ans;
int i, len = strlen(str);
for (i = 0; i < len; i += 4) {
(*ret++) = table[str[i]] << 2 | table[str[i + 1]] >> 4;
if (str[i + 2] != '=')
(*ret++) = (table[str[i + 1]] & 0x0f) << 4 | table[str[i + 2]] >> 2;
if (str[i + 3] != '=')
(*ret++) = table[str[i + 2]] << 6 | table[str[i + 3]];
}
}
int main()
{
init();
printf("%d\n",(int)table[0]);
scanf("%s",str);
decode(str);
printf("%s\n",ans)
return 0;
}输出的第二行一定是由小写字母、大写字母、数字和“+”、“/”、“=”构成的字符串。
#include <stdio.h>
#include <string.h>
char base[64];
char table[256];
char str[256];
char ans[256];
void init()
{
for (int i = 0; i < 26; i++) base[i] = 'A' + i;
for (int i = 0; i < 26; i++) base[26 + i] = 'a' + i;
for (int i = 0; i < 10; i++) base[52 + i] = '0' + i;
base[62] = '+', base[63] = '/';
for (int i = 0; i < 256; i++) table[i] = 0xff;
for (int i = 0; i < 64; i++) table[base[i]] = i;
table['='] = 0;
}
void decode(char *str)
{
char *ret = ans;
int i, len = strlen(str);
for (i = 0; i < len; i += 4) {
(*ret++) = table[str[i]] << 2 | table[str[i + 1]] >> 4;
if (str[i + 2] != '=')
(*ret++) = (table[str[i + 1]] & 0x0f) << 4 | table[str[i + 2]] >> 2;
if (str[i + 3] != '=')
(*ret++) = table[str[i + 2]] << 6 | table[str[i + 3]];
}
}
int main()
{
init();
printf("%d\n",(int)table[0]);
scanf("%s",str);
decode(str);
printf("%s\n",ans)
return 0;
}可能存在输入不同,但输出的第二行相同的情形。
#include <stdio.h>
#include <string.h>
char base[64];
char table[256];
char str[256];
char ans[256];
void init()
{
for (int i = 0; i < 26; i++) base[i] = 'A' + i;
for (int i = 0; i < 26; i++) base[26 + i] = 'a' + i;
for (int i = 0; i < 10; i++) base[52 + i] = '0' + i;
base[62] = '+', base[63] = '/';
for (int i = 0; i < 256; i++) table[i] = 0xff;
for (int i = 0; i < 64; i++) table[base[i]] = i;
table['='] = 0;
}
void decode(char *str)
{
char *ret = ans;
int i, len = strlen(str);
for (i = 0; i < len; i += 4) {
(*ret++) = table[str[i]] << 2 | table[str[i + 1]] >> 4;
if (str[i + 2] != '=')
(*ret++) = (table[str[i + 1]] & 0x0f) << 4 | table[str[i + 2]] >> 2;
if (str[i + 3] != '=')
(*ret++) = table[str[i + 2]] << 6 | table[str[i + 3]];
}
}
int main()
{
init();
printf("%d\n",(int)table[0]);
scanf("%s",str);
decode(str);
printf("%s\n",ans)
return 0;
}输出的第一行为“-1”。
#include <stdio.h>
#include <string.h>
char base[64];
char table[256];
char str[256];
char ans[256];
void init()
{
for (int i = 0; i < 26; i++) base[i] = 'A' + i;
for (int i = 0; i < 26; i++) base[26 + i] = 'a' + i;
for (int i = 0; i < 10; i++) base[52 + i] = '0' + i;
base[62] = '+', base[63] = '/';
for (int i = 0; i < 256; i++) table[i] = 0xff;
for (int i = 0; i < 64; i++) table[base[i]] = i;
table['='] = 0;
}
void decode(char *str)
{
char *ret = ans;
int i, len = strlen(str);
for (i = 0; i < len; i += 4) {
(*ret++) = table[str[i]] << 2 | table[str[i + 1]] >> 4;
if (str[i + 2] != '=')
(*ret++) = (table[str[i + 1]] & 0x0f) << 4 | table[str[i + 2]] >> 2;
if (str[i + 3] != '=')
(*ret++) = table[str[i + 2]] << 6 | table[str[i + 3]];
}
}
int main()
{
init();
printf("%d\n",(int)table[0]);
scanf("%s",str);
decode(str);
printf("%s\n",ans)
return 0;
}设输入字符串长度为n,decode函数的时间复杂度为( )。
#include <stdio.h>
#include <string.h>
char base[64];
char table[256];
char str[256];
char ans[256];
void init()
{
for (int i = 0; i < 26; i++) base[i] = 'A' + i;
for (int i = 0; i < 26; i++) base[26 + i] = 'a' + i;
for (int i = 0; i < 10; i++) base[52 + i] = '0' + i;
base[62] = '+', base[63] = '/';
for (int i = 0; i < 256; i++) table[i] = 0xff;
for (int i = 0; i < 64; i++) table[base[i]] = i;
table['='] = 0;
}
void decode(char *str)
{
char *ret = ans;
int i, len = strlen(str);
for (i = 0; i < len; i += 4) {
(*ret++) = table[str[i]] << 2 | table[str[i + 1]] >> 4;
if (str[i + 2] != '=')
(*ret++) = (table[str[i + 1]] & 0x0f) << 4 | table[str[i + 2]] >> 2;
if (str[i + 3] != '=')
(*ret++) = table[str[i + 2]] << 6 | table[str[i + 3]];
}
}
int main()
{
init();
printf("%d\n",(int)table[0]);
scanf("%s",str);
decode(str);
printf("%s\n",ans)
return 0;
}当输入为“Y3Nx”时,输出的第二行为( )。
#include <stdio.h>
#include <string.h>
char base[64];
char table[256];
char str[256];
char ans[256];
void init()
{
for (int i = 0; i < 26; i++) base[i] = 'A' + i;
for (int i = 0; i < 26; i++) base[26 + i] = 'a' + i;
for (int i = 0; i < 10; i++) base[52 + i] = '0' + i;
base[62] = '+', base[63] = '/';
for (int i = 0; i < 256; i++) table[i] = 0xff;
for (int i = 0; i < 64; i++) table[base[i]] = i;
table['='] = 0;
}
void decode(char *str)
{
char *ret = ans;
int i, len = strlen(str);
for (i = 0; i < len; i += 4) {
(*ret++) = table[str[i]] << 2 | table[str[i + 1]] >> 4;
if (str[i + 2] != '=')
(*ret++) = (table[str[i + 1]] & 0x0f) << 4 | table[str[i + 2]] >> 2;
if (str[i + 3] != '=')
(*ret++) = table[str[i + 2]] << 6 | table[str[i + 3]];
}
}
int main()
{
init();
printf("%d\n",(int)table[0]);
scanf("%s",str);
decode(str);
printf("%s\n",ans)
return 0;
}当输入为“Y2NmIDIwMjE=”时,输出的第二行为( )。
#include <stdio.h>
#define n 100000
#define N n + 1
int m;
int a[N], b[N], c[N], d[N];
int f[n], g[N];
void init()
{
f[1] = g[1] = 1;
for (int i = 2; i <= n; i++) {
if (!a[i]) {
b[m++] = i;
c[i] = 1, f[i] = 2;
d[i] = 1, g[i] = i + 1;
}
for (int j = 0; j < m && b[j] * i <= n; j++) {
int k = b[j];
a[i * k] = 1;
if (i % k == 0) {
c[i * k] = c[i] + 1;
f[i * k] = f[i] / c[i * k] * (c[i * k] + 1);
d[i * k] = d[i];
g[i * k] = g[i] * k + d[i];
break;
}
else {
c[i * k] = 1;
f[i * k] = 2 * f[i];
d[i * k] = g[i];
g[i * k] = g[i] * (k + 1);
}
}
}
}
int main()
{
init();
int x;
scanf("%d", &x);
printf("%d %d\n", f[x], g[x]);
return 0;
}假设输入的x是不超过1000的自然数,若输入不为"1",把第12行删去不会影响输出的结果。
#include <stdio.h>
#define n 100000
#define N n + 1
int m;
int a[N], b[N], c[N], d[N];
int f[n], g[N];
void init()
{
f[1] = g[1] = 1;
for (int i = 2; i <= n; i++) {
if (!a[i]) {
b[m++] = i;
c[i] = 1, f[i] = 2;
d[i] = 1, g[i] = i + 1;
}
for (int j = 0; j < m && b[j] * i <= n; j++) {
int k = b[j];
a[i * k] = 1;
if (i % k == 0) {
c[i * k] = c[i] + 1;
f[i * k] = f[i] / c[i * k] * (c[i * k] + 1);
d[i * k] = d[i];
g[i * k] = g[i] * k + d[i];
break;
}
else {
c[i * k] = 1;
f[i * k] = 2 * f[i];
d[i * k] = g[i];
g[i * k] = g[i] * (k + 1);
}
}
}
}
int main()
{
init();
int x;
scanf("%d", &x);
printf("%d %d\n", f[x], g[x]);
return 0;
}假设输入的x是不超过1000的自然数,第24行的"f[i]/c[i*k]"可能存在无法整除而向下取整的情况。
#include <stdio.h>
#define n 100000
#define N n + 1
int m;
int a[N], b[N], c[N], d[N];
int f[n], g[N];
void init()
{
f[1] = g[1] = 1;
for (int i = 2; i <= n; i++) {
if (!a[i]) {
b[m++] = i;
c[i] = 1, f[i] = 2;
d[i] = 1, g[i] = i + 1;
}
for (int j = 0; j < m && b[j] * i <= n; j++) {
int k = b[j];
a[i * k] = 1;
if (i % k == 0) {
c[i * k] = c[i] + 1;
f[i * k] = f[i] / c[i * k] * (c[i * k] + 1);
d[i * k] = d[i];
g[i * k] = g[i] * k + d[i];
break;
}
else {
c[i * k] = 1;
f[i * k] = 2 * f[i];
d[i * k] = g[i];
g[i * k] = g[i] * (k + 1);
}
}
}
}
int main()
{
init();
int x;
scanf("%d", &x);
printf("%d %d\n", f[x], g[x]);
return 0;
}假设输入的x是不超过1000的自然数,在执行完init()后,f数组不是单调递增的,但g数组是单调递增的。
#include <stdio.h>
#define n 100000
#define N n + 1
int m;
int a[N], b[N], c[N], d[N];
int f[n], g[N];
void init()
{
f[1] = g[1] = 1;
for (int i = 2; i <= n; i++) {
if (!a[i]) {
b[m++] = i;
c[i] = 1, f[i] = 2;
d[i] = 1, g[i] = i + 1;
}
for (int j = 0; j < m && b[j] * i <= n; j++) {
int k = b[j];
a[i * k] = 1;
if (i % k == 0) {
c[i * k] = c[i] + 1;
f[i * k] = f[i] / c[i * k] * (c[i * k] + 1);
d[i * k] = d[i];
g[i * k] = g[i] * k + d[i];
break;
}
else {
c[i * k] = 1;
f[i * k] = 2 * f[i];
d[i * k] = g[i];
g[i * k] = g[i] * (k + 1);
}
}
}
}
int main()
{
init();
int x;
scanf("%d", &x);
printf("%d %d\n", f[x], g[x]);
return 0;
}假设输入的x是不超过1000的自然数,init函数的时间复杂度为( )。
#include <stdio.h>
#define n 100000
#define N n + 1
int m;
int a[N], b[N], c[N], d[N];
int f[n], g[N];
void init()
{
f[1] = g[1] = 1;
for (int i = 2; i <= n; i++) {
if (!a[i]) {
b[m++] = i;
c[i] = 1, f[i] = 2;
d[i] = 1, g[i] = i + 1;
}
for (int j = 0; j < m && b[j] * i <= n; j++) {
int k = b[j];
a[i * k] = 1;
if (i % k == 0) {
c[i * k] = c[i] + 1;
f[i * k] = f[i] / c[i * k] * (c[i * k] + 1);
d[i * k] = d[i];
g[i * k] = g[i] * k + d[i];
break;
}
else {
c[i * k] = 1;
f[i * k] = 2 * f[i];
d[i * k] = g[i];
g[i * k] = g[i] * (k + 1);
}
}
}
}
int main()
{
init();
int x;
scanf("%d", &x);
printf("%d %d\n", f[x], g[x]);
return 0;
}假设输入的x是不超过1000的自然数,在执行完init()后,f[1], f[2], f[3] ...... f[100]中有( )个等于2。
#include <stdio.h>
#define n 100000
#define N n + 1
int m;
int a[N], b[N], c[N], d[N];
int f[n], g[N];
void init()
{
f[1] = g[1] = 1;
for (int i = 2; i <= n; i++) {
if (!a[i]) {
b[m++] = i;
c[i] = 1, f[i] = 2;
d[i] = 1, g[i] = i + 1;
}
for (int j = 0; j < m && b[j] * i <= n; j++) {
int k = b[j];
a[i * k] = 1;
if (i % k == 0) {
c[i * k] = c[i] + 1;
f[i * k] = f[i] / c[i * k] * (c[i * k] + 1);
d[i * k] = d[i];
g[i * k] = g[i] * k + d[i];
break;
}
else {
c[i * k] = 1;
f[i * k] = 2 * f[i];
d[i * k] = g[i];
g[i * k] = g[i] * (k + 1);
}
}
}
}
int main()
{
init();
int x;
scanf("%d", &x);
printf("%d %d\n", f[x], g[x]);
return 0;
}假设输入的x是不超过1000的自然数,当输入"1000"时,输出为( )。