http://acm.hdu.edu.cn/contests/contest_show.php?cid=802
三人三台电脑比赛，然而有个队友鸽了,只有两人打，一共过了5题
1001：水题，只需判断3和4的倍数即可，solved by lyy
#include <bits/stdc++.h>
using namespace std;
#define ll long long
int t,n;
int main()
{
scanf("%d",&t);
while (t--)
{
scanf("%d",&n);
if (n%3==0) printf("%lld\n",(ll)(n/3)*(n/3)*(n/3));
else if (n%4==0) printf("%lld\n",2LL*(n/4)*(n/4)*(n/4));
else printf("-1\n");
}
return 0;
}1003：水题，排序之后顺序取点即可, solved by lyy
#include <bits/stdc++.h>
using namespace std;
#define ll long long
int t,n;
struct point
{
int id;
int x,y;
}p[3005];
bool cmp(point a,point b)
{
return a.x<b.x;
}
int main()
{
scanf("%d",&t);
while (t--)
{
scanf("%d",&n);
for (int i=0;i<3*n;i++)
{
scanf("%d%d",&p[i].x,&p[i].y);
p[i].id=i+1;
}
sort(p,p+n*3,cmp);
for (int i=0;i<n;i++)
{
printf("%d %d %d\n",p[3*i].id,p[3*i+1].id,p[3*i+2].id);
}
}
return 0;
}1011：字符串处理，solved by sdn
/*
ID: oodt
PROG:
LANG:C++
*/
#include<iostream>
#include<algorithm>
#include<cstdio>
#include<cmath>
#include<string>
#include<cstring>
#include<map>
#include<vector>
#include<queue>
#include<stack>
#include<set>
#include<sstream>
using namespace std;
const int maxx=10005;
int n,m,k;
int a[maxx];
int ans = 0,cnt = 0,pos = 0;
int l = 0,r = 0;
int main()
{
#ifdef LOCAL
// freopen("","r",stdin);
#endif
int T;
scanf("%d",&T);
getchar();
char str[100];
int h,m,x,y;
while(T--)
{
scanf("%d%d%s",&h,&m,str);
if(str[3] == '+')
{
if(str[5] <= '9' && str[5] >= '0') x = 10*(str[4]-'0') + str[5] - '0';
else x = str[4] - '0';
if(str[5] == '.' || str[6] == '.')
{
if(str[5] == '.')
y = str[6]-'0';
else
y = str[7]-'0';
h += x-8;
m += y * 6;
}
else {
h += x-8;
}
}
else {
if(str[5] <= '9' && str[5] >= '0') x = 10*(str[4]-'0') + str[5] - '0';
else x = str[4] - '0';
if(str[5] == '.' || str[6] == '.')
{
if(str[5] == '.')
y = str[6]-'0';
else
y = str[7]-'0';
h += -x-8;
m -= y * 6;
}
else {
h += -8-x;
}
}
if(m < 0) {
m = (m + 60) % 60;
h--;
}
if(m >= 60){
m = m % 60;
h++;
}
if(h < 0) h = (h + 24)% 24;
if(h >= 24) h = h % 24;
printf("%02d:%02d\n",h,m);
}
return 0;
}1004：排序之后，直接暴力往数组里填数即可， solved by lyy
#include <bits/stdc++.h>
using namespace std;
#define ll long long
int t,n,m;
int a[100005];
int sum[100005];
struct st
{
int l,r;
}p[100005];
bool cmp(st a,st b)
{
if (a.l!=b.l) return a.l<b.l;
else return a.r<b.r;
}
int main()
{
scanf("%d",&t);
while (t--)
{
memset(a,0,sizeof(a));
memset(sum,0,sizeof(sum));
scanf("%d%d",&n,&m);
for (int i=1;i<=m;i++)
{
scanf("%d%d",&p[i].l,&p[i].r);
}
sort(p+1,p+m+1,cmp);
int l,r;
for (int i=1;i<p[1].l;i++)
{
a[i]=1;
}
for (int i=p[1].l;i<=p[1].r;i++)
{
a[i]=1-p[1].l+i;
sum[a[i]]=1;
}
l=p[1].l;r=p[1].r;
int mi=a[p[1].r]+1;
for (int i=2;i<=m;i++)
{
if (p[i].l>r)
{
for (int j=l;j<=r;j++)
{
sum[a[j]]=0;
l++;
}
for (int j=l;j<p[i].l;j++)
{
a[j]=1;
}
l=p[i].l;
r=p[i].l-1;
for (int j=p[i].l;j<=p[i].r;j++)
{
a[j]=1-p[i].l+j;
sum[a[j]]=1;
r++;
}
mi=p[i].r-p[i].l+2;
}
else
{
for (int j=l;j<p[i].l;j++)
{
sum[a[j]]=0;
mi=min(mi,a[j]);
l++;
}
if (p[i].r>r)
{
for (int j=r;j<p[i].r;j++)
{
r++;
a[r]=mi;
sum[a[r]]=1;
while (sum[mi]>0) mi++;
}
}
}
}
int ma=p[1].r;
for (int i=1;i<=m;i++)
{
ma=max(ma,p[i].r);
}
for (int i=ma+1;i<=n;i++)
{
a[i]=1;
}
for (int i=1;i<=n;i++)
{
printf("%d",a[i]);
if (i!=n) printf(" ");
}
printf("\n");
}
return 0;
}1007：打表，然后扔到OEIS里，找到一段python代码，研究一下发现可以二分写, solved by lyy
扫描二维码关注公众号，回复：
2468803 查看本文章
#include <bits/stdc++.h>
using namespace std;
#define ll long long
const ll mod=1000000007;
int t;
ll n;
long long inv(long long a,long long m)
{
if(a == 1)return 1;
return inv(m%a,m)*(m-m/a)%m;
}
ll inv2=inv(2,mod);
ll f(ll x)
{
if (x==1) return 1;
else return x+f(x/2);
}
ll find(ll n)
{
ll l=1,r=n;
ll m;
while (l<=r)
{
m=(l+r)/2;
if (f(m)<n) l=m+1;
else r=m-1;
}
return l;
}
ll ff(ll n)
{
if (n==1) return 1;
ll ans=(n%mod)*((n+1)%mod)%mod*inv2%mod;
return (ans+2*ff(n/2))%mod;
}
int main()
{
scanf("%d",&t);
while (t--)
{
scanf("%lld",&n);
if (n<3) printf("%d\n",n);
else
{
ll p=find(n-1);
ll ans=0;
ans+=p%mod*(n-1-f(p-1))%mod;
ans=(ans+mod)%mod;
ans=(ans+ff(p-1)+1)%mod;
printf("%lld\n",ans);
}
}
return 0;
}