Day 29 算法笔记之数学问题 5.1简单数学(2)

1.数列的片段和

找到关系很重要

#include <cstdio>
#include <cctype>
#include <cstring>
#include <math.h>
#include <iostream>
#include <algorithm>
#include <map>
#include <sys/time.h>
using namespace std;

int main(){
	
	int n;
	double v,ans=0;
	scanf("%d",&n);
	for(int i=0;i<n;i++){
		scanf("%lf",&v);
		ans+=v*i*(n+1-i);
	}
	printf("%.2f\n",ans);
	
	return 0;
} 

2.Elevator

#include <cstdio>
#include <cctype>
#include <cstring>
#include <math.h>
#include <iostream>
#include <algorithm>
#include <map>
#include <sys/time.h>
using namespace std;

int main(){
	
	int n;
	int martix[100];
	scanf("%d",&n);
	for(int i=0;i<n;i++){
		scanf("%d",&martix[i]);
	}
	
	int now=0,title=0;
	for(int i=0;i<n;i++){
		if(martix[i]==now){
			title+=5;
		}else if(martix[i]>now){
			title+=6*(martix[i]-now)+5;
		}else{
			title+=4*(now-martix[i])+5;
		}
		now = martix[i];
		printf("%d\n",title);
	}
	printf("%d\n",title);
	return 0;
} 

3.counting ones

#include <cstdio>
#include <cctype>
#include <cstring>
#include <math.h>
#include <iostream>
#include <algorithm>
#include <map>
#include <sys/time.h>
using namespace std;

int main(){
	
	int n,a=1,ans=0;
	int left,now,right;
	scanf("%d",&n);
	
	while(n/a!=0){
		left = n/(a*10);
		now = n/a%10;
		right = n%a;
		if(now==0) ans+=left*a;
		else if(now==1) ans+=left*a+right+1;
		else ans+=(left+1)*a;
		a*=10;
	}
	printf("%d\n",ans);
	return 0;
	
	return 0;
}

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转载自blog.csdn.net/aixiaoxiao13/article/details/120913170
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