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;
}