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Codeforces Global Round 19
2022-07-06 03:35:00 【Changersh】
A. Sorting Parts
Sign in problem
Ideas
n Number , Separately 1~(n-1) Length before and after the infix to sort , If a certain time is not in non descending order, output "YES", Otherwise output "NO"
It's because the length of the prefix is from 1 Start , So as long as there is a number in the original array, it is not in non descending order , Then the non descending order must not be satisfied in a subsequent sort
eg: 2 1 4 5 6
Sort , First of all, yes The length is 1 Pre suffix sort of , Still 2 1 4 5 6 Does not satisfy non descending
Code
#include<stdio.h>
#include<string.h>
#include<stdlib.h>
typedef int Bool;
typedef long long ll;
#define MS(a, b) memset(a, b, sizeof(a))
int compare(const void* a, const void* b);
ll s[N];
int main() {
int T;
scanf("%d", &T);
while (T--) {
int n = 0;
MS(s, 0);
scanf("%d", &n);
for (int i = 0; i < n; i++) {
scanf("%lld", &s[i]);
}
int flag = 0;
for (int i = 1; i < n; i++) {
if (s[i] < s[i - 1]) flag = 1;
}
if (flag) printf("YES\n");
else printf("NO\n");
}
return 0;
}
int compare(const void* a, const void* b) {
ElementType * s1 = (ElementType*)a;
ElementType * s2 = (ElementType*)b;
if (*s1 > *s2) return 1;
else
if (*s1 == *s2) return 0;
else return -1;
}
C. Andrew and Stones
Sign in problem … But I still don't understand
thinking road
n Rubble , Every pile ai individual , Choose three numbers 1 ≤ i < j < k ≤ n, also j Must be greater than or equal to 2
take j Medium stone , Each direction i Let's play one. towards k Let's play one.
Ask if you can move all the stones to In the first pile and the last pile
Sub situation
- n == 3 , If the middle is an odd number, it cannot
- n For any , The middle is full of 1, Do not move stones , Output -1
- Satisfied
So directly judge whether it is all 1 that will do ,
Then record the minimum number of steps as ,(s[i] + 1) / 2 And
Code
#include<stdio.h>
#include<stdlib.h>
#include<string.h>
#include<math.h>
#include<time.h>
typedef int Bool;
typedef long long ll;
typedef unsigned long long ull;
#define N 100000+5
#define MS(a, b) memset(a, b, sizeof(a))
/* Except for n == 3 The number in the middle of time is odd , And arbitrary n All medians of are less than 2, Anything else is ok */
ll s[N], sum, ans;
int main() {
int T, n;
scanf("%d", &T);
while (T--) {
sum = 0;// Record the total number of steps (s[i] + 1) / 2
ans = 0;// Judge whether it's all 1
MS(s, 0);
scanf("%d", &n);
for (int i = 0; i < n; i++)
scanf("%lld", &s[i]);
for (int i = 1; i < n - 1; i++) {
ans |= (s[i] > 1) ;
sum += (s[i] + 1) / 2;
}
if (!ans || (n == 3 && s[1] % 2 == 1)) printf("-1\n");
else printf("%lld\n", sum);
}
return 0;
}
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