1205A Almost Equal
找规律，n为奇数的时候可以，n为偶数的时候不行，分析一下样例每个元素的摆放位置就能找到规律。
#include <stdio.h>
#include <time.h>
#include <string.h>
#include <algorithm>
#include <stack>
#include <vector>
#include <set>
#include <iostream>
#include <queue>
#include <string>
#include <math.h>
typedef long long LL;
using namespace std;
const int inf = 0x3f3f3f3f;
const int mod = 1e9 + 7;
#define mid(l,r) (( l + r ) / 2)
#define lowbit(x) (( x & ( - x )))
#define lc(root) ((root * 2))
#define rc(root) ((root * 2 + 1))
#define me(array,x) (memset( array , x , sizeof( array ) ) )
const int maxn = 1e6;
LL a[maxn];
int main()
{
int n,m,flag = 0;
scanf("%d",&n);
if(n%2==0)flag = 1;
else {
int s = 2;
a[1] = 1 , a[2 * n] = 2 * n;
for(int i = 2 , j = n + 1 ; i <= n ; i += 2 , j += 2){
a[j] = s++;
a[j+1] = s++;
a[i] = s++;
a[i+1] = s++;
}
}
if(flag )printf("NO\n");
else {
printf("YES\n");
for(int i=1;i <= 2*n;i++)
printf("%d\n",a[i]);
}
return 0;
}
1206A Choose Two Numbers
找最大的两个数就行了
#include <stdio.h>
#include <time.h>
#include <string.h>
#include <algorithm>
#include <stack>
#include <vector>
#include <set>
#include <iostream>
#include <queue>
#include <string>
#include <math.h>
typedef long long ll;
using namespace std;
const int inf = 0x3f3f3f3f;
const int mod = 1e9 + 7;
#define mid(l,r) (( l + r ) / 2)
#define lowbit(x) (( x & ( - x )))
#define lc(root) ((root * 2))
#define rc(root) ((root * 2 + 1))
#define me(array,x) (memset( array , x , sizeof( array ) ) )
const int maxn = 1e3;
int a[maxn],b[maxn];
int main()
{
int n,m;
scanf("%d",&n);
for(int i = 0 ; i < n ; i++)scanf("%d",&a[i]);
scanf("%d",&m);
for(int i = 0 ; i < m ; i++)scanf("%d",&b[i]);
sort(a,a+n);sort(b,b+m);
printf("%d %d\n",a[n-1],b[m-1]);
return 0;
}
1206B Make Product Equal One
负数就直接变成-1，正数就变成1，注意0的个数，如果没有0且负数有奇数个，那就还需要把一个负数变成1
#include <stdio.h>
#include <time.h>
#include <string.h>
#include <algorithm>
#include <stack>
#include <vector>
#include <set>
#include <iostream>
#include <queue>
#include <string>
#include <math.h>
typedef long long LL;
using namespace std;
const int inf = 0x3f3f3f3f;
const int mod = 1e9 + 7;
#define mid(l,r) (( l + r ) / 2)
#define lowbit(x) (( x & ( - x )))
#define lc(root) ((root * 2))
#define rc(root) ((root * 2 + 1))
#define me(array,x) (memset( array , x , sizeof( array ) ) )
const int maxn = 1e6;
LL a[maxn];
int main()
{
int n,m;
scanf("%d",&n);
LL ans = 0;
int flag1 = 0 , flag2 = 0 ;
for(int i = 1 ; i <= n ; i++){
scanf("%lld",&a[i]);
if(a[i] < 0 )flag1++;
if(a[i] == 0)flag2++;
}
for(int i = 1 ; i <= n ; i++){
LL x = a[i];
if(x < 0)ans += ((-x)-1);
if(x == 0)ans += 1;
if(x > 0)ans += (x-1);
}
if(flag1 % 2 == 1 && flag2 == 0)ans += 2;
printf("%lld\n",ans);
return 0;
}
B题第一次提交过了，手贱改了一下再交一次，还迷迷糊糊的还把pretest过了，最后主测试的时候就直接跳过了第一次的提交，然后第二次交的就wa了。。。。结束之后就试了一下第一次提交的对不对，结果是正确的。。。本来是可以出三个题的
