LevOJ.sln - 第三期
LevOJ平台.sln
P2097 乘法口诀数列
问题描述
解决方法
#include <stdio.h>
int main() {
int ans[100];
int a1,a2,n;
scanf("%d%d%d",&a1,&a2,&n);
ans[0]=a1;
ans[1]=a2;
int j=2;
for(int i=0;i<10;i++){
if (ans[i]*ans[i+1]>9)
{
ans[j]=(ans[i]*ans[i+1])/10;
j++;
ans[j]=(ans[i]*ans[i+1])%10;
j++;
}
else
{
ans[j]=ans[i]*ans[i+1];
j++;
}
}
for(int i=0;i<n;i++){
printf("%d ",ans[i]);
}
printf("\n");
return 0;
}
P2098 试试手气
问题描述
解决方法
#include <stdio.h>
int main()
{
int index[6][6] = {0};
int n;
for (int i = 0; i < 6; i++)
{
scanf("%d", &index[0][i]);
}
scanf("%d", &n);
for (int i = 0; i < 6; i++)
{
int temp = 6;
for (int j = 1; j <= 5;)
{
if (temp==index[0][i])
{
temp--;
}
else
{
index[j][i] = temp;
temp--;
j++;
}
}
}
for (int i = 0; i < 6; i++)
{
printf("%d ", index[n][i]);
}
printf("\n");
return 0;
}
P2099 机工士姆斯塔迪奥
问题描述
解决方法
#include <stdio.h>
int main() {
int n, m, s, map[100][100]={0};
scanf("%d%d%d", &n, &m, &s);
int rc, ind, cnt=0;
for(int i=0; i<s; i++) {
scanf("%d%d", &rc, &ind);
switch (rc)
{
case 0:
for (int i = 0; i < m; i++)
{
map[ind-1][i] = 1;
}
break;
case 1:
for (int i = 0; i < n; i++)
{
map[i][ind-1] = 1;
}
break;
default:
break;
}
}
for(int i=0; i<n; i++) {
for(int j=0; j<m; j++) {
if(map[i][j] == 0) {
cnt++;
}
}
}
printf("%d", cnt);
return 0;
}
P1992 猜帽子游戏
问题描述
输入输出样例
解决方法
#include <stdio.h>
int main()
{
int n;
scanf("%d", &n);
int hats[100];
for (int i = 0; i < n; i++)
scanf("%d", &hats[i]);
int k;
scanf("%d", &k);
for (int i = 0; i < k; i++)
{
int guesses[100];
int ycnt = 0, ncnt = 0, hcnt = 0;
for (int j = 0; j < n; j++)
{
scanf("%d", &guesses[j]);
if (guesses[j] != 0)
{
hcnt = 1;
if (guesses[j] == hats[j])
{
ycnt++;
}
else
{
ncnt++;
}
}
}
if (ncnt == 0 && ycnt > 0 && hcnt)
{
printf("Da Jiang!!!\n");
}
else
{
printf("Ai Ya\n");
}
}
return 0;
}
P2023 超人与奥特曼
问题描述
解决方法
#include <stdio.h>
enum
{
M = 100
};
int main()
{
int x[M];
int aotoaman = 0, surperman = 0;
int n, k = 0;
scanf("%d", &n);
int l = n - 1;
for (int i = 0; i < n; i++)
{
scanf("%d", &x[i]);
}
while (k != l + 1)
{
if (aotoaman >= surperman)
{
surperman += x[k++];
}
else if (aotoaman < surperman)
{
aotoaman += x[l--];
}
}
printf("%d %d", surperman, aotoaman);
return 0;
}
P1296 分形宇宙
问题描述
解决方法
#include <stdio.h>
#include <math.h>
char canvas[10000][10000];
void initCanvas(int size)
{
for (int i = 0; i < size; i++)
{
for (int j = 0; j < size; j++)
{
canvas[i][j] = ' ';
}
}
}
void fra(int n, int x, int y)
{
int size = pow(3, n - 2);
if (n == 1)
{
canvas[x][y] = 'X';
return;
}
fra(n - 1, x, y);
fra(n - 1, x + 2 * size, y);
fra(n - 1, x + size, y + size);
fra(n - 1, x, y + 2 * size);
fra(n - 1, x + 2 * size, y + 2 * size);
}
int main()
{
int n;
scanf("%d", &n);
int size = pow(3, n - 1);
initCanvas(size);
fra(n, 0, 0);
for (int i = 0; i < size; i++)
{
for (int j = 0; j < size; j++)
{
printf("%c", canvas[i][j]);
}
printf("\n");
}
return 0;
}
这是一道非常有趣的题 可玩性非常高~
有关分形，看看这篇介绍
有关本题，可以看看这篇思路讲解
另 （不建议使用的）偷鸡法
点击展开代码
#include <stdio.h>
int main()
{
int n;
scanf("%d", &n);
switch (n)
{
case 1:
printf(
"X\n");
break;
case 2:
printf(
"X X\n"
" X \n"
"X X\n");
break;
case 3:
printf(
"X X X X\n"
" X X \n"
"X X X X\n"
" X X \n"
" X \n"
" X X \n"
"X X X X\n"
" X X \n"
"X X X X\n");
break;
case 4:
printf(
"X X X X X X X X\n"
" X X X X \n"
"X X X X X X X X\n"
" X X X X \n"
" X X \n"
" X X X X \n"
"X X X X X X X X\n"
" X X X X \n"
"X X X X X X X X\n"
" X X X X \n"
" X X \n"
" X X X X \n"
" X X \n"
" X \n"
" X X \n"
" X X X X \n"
" X X \n"
" X X X X \n"
"X X X X X X X X\n"
" X X X X \n"
"X X X X X X X X\n"
" X X X X \n"
" X X \n"
" X X X X \n"
"X X X X X X X X\n"
" X X X X \n"
"X X X X X X X X\n");
break;
case 5:
printf(
"X X X X X X X X X X X X X X X X\n"
" X X X X X X X X \n"
"X X X X X X X X X X X X X X X X\n"
" X X X X X X X X \n"
" X X X X \n"
" X X X X X X X X \n"
"X X X X X X X X X X X X X X X X\n"
" X X X X X X X X \n"
"X X X X X X X X X X X X X X X X\n"
" X X X X X X X X \n"
" X X X X \n"
" X X X X X X X X \n"
" X X X X \n"
" X X \n"
" X X X X \n"
" X X X X X X X X \n"
" X X X X \n"
" X X X X X X X X \n"
"X X X X X X X X X X X X X X X X\n"
" X X X X X X X X \n"
"X X X X X X X X X X X X X X X X\n"
" X X X X X X X X \n"
" X X X X \n"
" X X X X X X X X \n"
"X X X X X X X X X X X X X X X X\n"
" X X X X X X X X \n"
"X X X X X X X X X X X X X X X X\n"
" X X X X X X X X \n"
" X X X X \n"
" X X X X X X X X \n"
" X X X X \n"
" X X \n"
" X X X X \n"
" X X X X X X X X \n"
" X X X X \n"
" X X X X X X X X \n"
" X X X X \n"
" X X \n"
" X X X X \n"
" X X \n"
" X \n"
" X X \n"
" X X X X \n"
" X X \n"
" X X X X \n"
" X X X X X X X X \n"
" X X X X \n"
" X X X X X X X X \n"
" X X X X \n"
" X X \n"
" X X X X \n"
" X X X X X X X X \n"
" X X X X \n"
" X X X X X X X X \n"
"X X X X X X X X X X X X X X X X\n"
" X X X X X X X X \n"
"X X X X X X X X X X X X X X X X\n"
" X X X X X X X X \n"
" X X X X \n"
" X X X X X X X X \n"
"X X X X X X X X X X X X X X X X\n"
" X X X X X X X X \n"
"X X X X X X X X X X X X X X X X\n"
" X X X X X X X X \n"
" X X X X \n"
" X X X X X X X X \n"
" X X X X \n"
" X X \n"
" X X X X \n"
" X X X X X X X X \n"
" X X X X \n"
" X X X X X X X X \n"
"X X X X X X X X X X X X X X X X\n"
" X X X X X X X X \n"
"X X X X X X X X X X X X X X X X\n"
" X X X X X X X X \n"
" X X X X \n"
" X X X X X X X X \n"
"X X X X X X X X X X X X X X X X\n"
" X X X X X X X X \n"
"X X X X X X X X X X X X X X X X\n");
break;
default:
break;
}
return 0;
}
P2100 考试座位号
问题描述
解决方法
#include <stdio.h>
void printInfo(long long int info[][3], int m, int n)
{
for (int i = 0; i < n; i++)
if (m == (int)info[i][1])
printf("%lld %lld\n", info[i][0], info[i][2]);
}
int main()
{
int n;
scanf("%d", &n);
long long int info[10][3];
for (int i = 0; i < n; i++)
scanf("%lld %lld %lld", &info[i][0], &info[i][1], &info[i][2]);
int m;
scanf("%d", &m);
int query[10];
for (int i = 0; i < m; i++)
scanf("%d", &query[i]);
for (int i = 0; i < m; i++)
printInfo(info, query[i], n);
return 0;
}
P2101 一帮一
问题描述
解决方法
#include <stdio.h>
int main()
{
int n;
scanf("%d", &n);
int gender[50];
char name[50][20];
int isUsed[50] = {0};
for (int i = 0; i < n; i++)
{
scanf("%d", &gender[i]);
scanf("%s", name[i]);
}
for (int i = 0; i < n; i++)
{
if (isUsed[i])
{
continue;
}
for (int j = n - 1; j >= 0; j--)
{
if (i != j && !isUsed[j] && gender[i] != gender[j])
{
printf("%s %s\n", name[i], name[j]);
isUsed[i] = 1;
isUsed[j] = 1;
break;
}
}
}
return 0;
}
P1325 区间合并
问题描述
解决方法
#include <stdio.h>
void sort(int s[], int e[], int n)
{
for (int i = 0; i < n - 1; i++)
for (int j = 0; j < n - 1 - i; j++)
if (s[j] > s[j + 1])
{
int temps = s[j];
int tempe = e[j];
s[j] = s[j + 1];
e[j] = e[j + 1];
s[j + 1] = temps;
e[j + 1] = tempe;
}
}
void merge(int s[], int e[], int n)
{
int start = s[0], end = e[0];
for (int i = 1; i < n; i++)
{
if (end < s[i])
{
printf("%d %d\n", start, end);
start = s[i];
end = e[i];
}
else if (end < e[i])
{
end = e[i];
}
}
printf("%d %d\n", start, end);
}
int main()
{
int n;
scanf("%d", &n);
int s[n], e[n];
for (int i = 0; i < n; i++) scanf("%d %d", &s[i], &e[i]);
sort(s, e, n);
merge(s, e, n);
return 0;
}
P2045 数字蚯蚓
问题描述
输入输出样例
解决方法
#include <stdio.h>
#include <string.h>
int main()
{
char str[200], be[200], af[200];
fgets(str, 200, stdin);
int len = strlen(str);
int k;
scanf("%d", &k);
strcpy(af, str);
for (int i = 0; i < k; i++)
{
af[i] = af[k];
}
strcpy(be, str);
for (int i = k; i < len - 1; i++)
{
be[i] = be[k - 1];
}
printf("%s", be);
printf("%s", af);
return 0;
}
第三期完结！！！
看完了点个赞再走嘛
