1 条题解

  • 0
    @ 2024-12-24 9:59:30

    C :

    #include<stdio.h>
    int N,x,y,Sum=1,k;
    int main(){
    	scanf("%d%d",&N,&x);
    	for(int i=2;i<=N;i++){
    		scanf("%d",&y);
    		if(y>x){//升序k=1
    			if(k==2||k==0){
    				k=1;
    				Sum++;
    				x=y;
    			}
    			if(k==1)x=y;//取大
    		}
    		if(y<x){//降序k=2
    			if(k==1||k==0){
    				k=2;
    				Sum++;
    				x=y;
    			}
    			if(k==2)x=y;//取小
    		}
    	}
    	printf("%d",Sum);
    }
    

    C++ :

    #include <cstdio>
    #include <algorithm>
    #include <cstring>
    using namespace std;
    
    int N=0;
    int nDate_A[1000001]={0},nDate_B[1000001]={0};
    int nFlower[1000001]={0};
    int nLen_A=1,nLen_B=1,maxx=0;
    
    int main(){
    	scanf("%d",&N);
    	for(int i=1;i<=N;++i){
    		scanf("%d",&nFlower[i]);
    	}
    	nDate_A[nLen_A]=nFlower[1];
    	nDate_B[nLen_B]=nFlower[1];
    	for(int i=2;i<=N;++i){//第A种
    		if(nLen_A%2==1){
    			if(nFlower[i]>nDate_A[nLen_A]) nDate_A[++nLen_A]=nFlower[i];
    			else nDate_A[nLen_A]=nFlower[i];
    		}
    		else{
    			if(nFlower[i]<nDate_A[nLen_A]) nDate_A[++nLen_A]=nFlower[i];
    			else nDate_A[nLen_A]=nFlower[i];
    		}
    		
    		if(nLen_B%2==1){
    			if(nFlower[i]<nDate_B[nLen_B]) nDate_B[++nLen_B]=nFlower[i];
    			else nDate_B[nLen_B]=nFlower[i];
    		}
    		else{
    			if(nFlower[i]>nDate_B[nLen_B]) nDate_B[++nLen_B]=nFlower[i];
    			else nDate_B[nLen_B]=nFlower[i];
    		}
    	}
    	printf("%d",max(nLen_B,nLen_A));
    	fclose(stdin);
    	fclose(stdout);
    	return 0;
    }
    
    
    • 1

    信息

    ID
    1640
    时间
    1000ms
    内存
    128MiB
    难度
    (无)
    标签
    递交数
    0
    已通过
    0
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