2015多校8

Solved	Pro.ID	Title	Author	Source	(AC/Submit)Ratio
	5380	Travel with candy	SXYZ	2015 Multi-University Training Contest 8	(46/103)44.66%
	5381	The sum of gcd	SXYZ	2015 Multi-University Training Contest 8	(208/496)41.94%
	5382	GCD?LCM!	SXYZ	2015 Multi-University Training Contest 8	(73/139)52.52%
	5383	Yu-Gi-Oh!	SXYZ	2015 Multi-University Training Contest 8	(95/366)25.96%
	5384	Danganronpa	SXYZ	2015 Multi-University Training Contest 8	(278/506)54.94%
	5385	The path	SXYZ	2015 Multi-University Training Contest 8	(202/546)37.00%
	5386	Cover	SXYZ	2015 Multi-University Training Contest 8	(377/1092)34.52%
	5387	Clock	SXYZ	2015 Multi-University Training Contest 8	(325/469)69.30%
	5388	Geometer's Sketchpad	SXYZ	2015 Multi-University Training Contest 8	(21/58)36.21%
	5389	Zero Escape	SXYZ	2015 Multi-University Training Contest 8	(485/953)50.89%
	5390	tree	SXYZ	2015 Multi-University Training Contest 8	(28/160)17.50%

HDU 5381

给定一个数组，多次询问，求区间[l,r]内所有子区间的最大公约数之和。

以a[x]为右端点的gcd区间情况最多只有log(a[x])种，因为a[x]的因子只有sqrt(a[x])种，a[x]的因子的因子只有sqrt(sqrt(a[x]))种,....依次类推共有log(a[x])种

如下图x=7的情况

#include<iostream>
#include<cstdio>
#include<cmath>
#include<cstring>
using namespace std;
const int fen=12*3600;
int hh,mm,ss;
int a[4],b[4];
int run(int a,int b)
{
int ans=abs(a-b);
ans%=fen;
if(ans>fen/2)ans=fen-ans;
return ans;
}
int gcd(int a,int b)
{
if(a%b==0)return b;
return gcd(b,a%b);
}
void print(int x)
{
x*=360;
int g=gcd(x,fen);
int t1=x/g;
int t2=fen/g;
if(t2==1)printf("%d",t1);
else printf("%d/%d",t1,t2);
}
int main()
{
int T;
scanf("%d",&T);
char str[100];
while(T--)
{
scanf("%s",str);
sscanf(str,"%d:%d:%d",&hh,&mm,&ss);
if(hh>=12)hh-=12;
a[1]=hh*3600+60*mm+ss;
a[2]=12*(60*mm+ss);
a[3]=12*60*ss;
b[1]=run(a[1],a[2]);
b[2]=run(a[1],a[3]);
b[3]=run(a[2],a[3]);
for(int i=1;i<=3;i++)
{
print(b[i]);
printf(" ");
}
puts("");
}
return 0;
}

View Code