#include<iostream>
#include<cstdio>
using namespace std;
unsigned long long n;
int main(){
cin>>n;
for(unsigned long long x=9223372036854775808,a=64;a!=-1;x/=2,a-=1){
if(n/x!=0&&a!=1){
cout<<"2^"<<a-1<<"+";
}
if(a==1){
cout<<"2^0";
}
}
return 0;
}

#include <cstdio>
#include <algorithm>
#include <cstring>
typedef long long L;
using namespace std;
L a, b, c, _10[15], ans[15], f[15][15][15];
bool flag;
void get_f(){
_10[0] = 1;
for(int i = 1; i < 15; i++) _10[i] = _10[i-1]*10;
for(int i = 1; i < 10; i++) f[i][1][i] = 1;
for(int i = 1; i < 10; i++)
for(int j = 2; j < 15; j++){
f[i][j][1] = _10[j-2]*(j-1);
for(int k = 2; k < 10; k++)
f[i][j][k] = f[i][j][1];
}
for(int i = 1; i < 10; i++)
for(int j = 2; j < 15; j++){
f[i][j][i] += _10[j-1];
}
}
void get_all(){
int la = 0, lb = 0;
L ta = a-1, tb = b, all = 0;
while(ta){
la++, ta /= 10;
}
while(tb){
lb++, tb /= 10;
}
for(int i = la; i < lb; i++){
all += (_10[i]-_10[i-1])*i;
} all += (b-_10[lb-1]+1)*lb;
all -= (a-_10[la-1])*la;
ans[0] = all;
}
L d(L k, int i){
if(!k) return 0;
int fir, len = 0;
L t = k, res = 0;
while(t){
len++, t /= 10;
} fir = k / _10[len-1];
if(len == 1) return i <= k;
for(int j = 1; j < fir; j++){
res += f[j][len][i];
}
res += d(k%_10[len-1], i) + d(_10[len-1]-1, i) + (fir==i)*(k%_10[len-1]+1);
return res;
}
int main()
{
freopen("count2013.in", "r", stdin);
freopen("count2013.out", "w", stdout);
scanf("%lld %lld", &b, &c);
a = 1;
get_f();
get_all();
for(int i = 1; i < 10; i++){
ans[i] = d(b, i) - d(a-1, i);
ans[0] -= ans[i];
}
printf("%lld ", ans[c]);
return 0;
}

#include <iostream>
#include<cstdio>
using namespace std;
int main()
{
//freopen("mas.in","r",stdin);
//freopen("mas.out","w",stdout);
int a[100][100];
int i,j,n;
int max,point,flag=1,work=1;
cin>>n;
for(i=0;i<n;i++)
for(j=0;j<n;j++)
cin>>a[i][j];
for(i=0;i<n;i++)
{
max=a[i][0];
point=0;
for(j=1;j<n;j++)
{
if(max>a[i][j])
{
max=a[i][j];
point=j;
}
}
flag=1;
for(j=0;j<n;j++)
{
if(a[j][point]>max)
flag=0;
}
if(flag)
{
cout<<i+1<<" "<<point+1<<" "<<max<<endl;
return 0;
}
}
cout<<"no"<<endl;
return 0;
}

回复 @天下第一的吃��殿下 :
#include<iostream>
#include<cstdio>
using namespace std;
int main()
{
freopen("snake.in","r",stdin);
freopen("snake.out","w",stdout);
int a[102][102],d=1;
int n;
cin>>n;
for(int i=1,x=1,y=1;i<=n*n;i++)
{
a[x][y]=i;
if(d==1){
x=x+1;
if(y==1)
d=2;
else
d=4;
}
else if(d==2)
{
x=x-1;y=y+1;
if(y==n)d=1;
else
if(x==1) d=3;
}
else if(d==3)
{
y=y+1;
if(x==n) d=2;
else
d=4;
}
else if(d==4)
{
x=x+1;y=y-1;
if(x==n) d=3;
else if(y==1) d=1;
}
}
for(int i=1;i<=n;i++)
{
for(int k=1;k<=n;k++)
{
cout<<a[i][k]<<" ";
}
cout<<endl;
}
return 0;}

其实和"稳定婚姻"差不多,不过这题没有给出一个最大匹配,但只是问整个图中是否存在交错环.

None.............

过了两个点之后发现样例过不去了……

写了这么麻烦的代码居然一共只有3个点？？？？！！！！！