1 hdu 1032
#include <iostream>
using namespace std;
int len(int n)
{
int s = 1;
while (n!=1)
{
if (n&1) n = 3*n +1;
else n /= 2;
s++;
}
return s;
}
int main()
{
int i, j, n, l, m;
while (~scanf("%d %d", &i, &j))
{
printf("%d %d", i, j);
if (i>j) swap(i, j);
m = 0;
for (;i<=j;++i)
{
l = len(i);
if (m<l) m = l;
}
printf(" %d\n", m);
}
return 0;
}
2 hdu 1029 （鴿巢原理）
#include <iostream>
using namespace std;
int main()
{
int ans, n, c, num;
while (~scanf("%d", &n))
{
c = 0;
for (int i=1;i<=n;++i)
{
scanf("%d", &ans);
if (c==0)
{
num = ans;
c ++;
}
else if (ans == num) c ++;
else c --;
}
printf("%d\n", num);
}
return 0;
}
3 hdu 1033 題目看到蛋碎
#include <iostream>
using namespace std;
int f[4][2] = {0,10,10,0,0,-10,-10,0};
void move(int &x, int &y, char c, int &s)
{
int u;
switch (s)
{
case 0: if (c=='V') u = 0, s = 1; else u = 2, s = 3; break;
case 1: if (c=='V') u = 3, s = 2; else u = 1, s = 0; break;
case 2: if (c=='V') u = 2, s = 3; else u = 0, s = 1; break;
case 3: if (c=='V') u = 1, s = 0; else u = 3, s = 2; break;
}
x += f[u][0];
y += f[u][1];
}
int main()
{
char c[300];
int i, k;
while (~scanf("%s", c))
{
k = strlen(c);
printf("300 420 moveto\n310 420 lineto\n");
int x = 310, y = 420, s = 0;
for (i=0;i<k;++i)
{
move(x, y, c[i], s);
printf("%d %d lineto\n", x, y);
}
printf("stroke\nshowpage\n");
}
return 0;
}
4 hdu 1036 sscanf的用法
#include <iostream>
using namespace std;
int main()
{
int f, n, i, a, sum, h, m, s;
double d;
char c[20], t[100];
scanf("%d%lf", &n, &d);
while (~scanf("%d", &a))
{
sum = f = 0;
for (i=1;i<=n;++i)
{
scanf("%s", c);
if (c[0] == '-') f = 1;
sscanf(c, "%d:%d:%d", &h, &m, &s);
sum += h*3600 + m*60 + s;
}
if (f) printf("%3d: -\n", a);
else
{
sum = int(sum/d + 0.5);
printf("%3d: %d:%2.2d min/km\n", a, sum/60, sum%60);
}
}
return 0;
}
5 hdu 1037 不解釋
#include <iostream>
using namespace std;
int main()
{
int h1, h2, h3;
while (~scanf("%d %d %d", &h1, &h2, &h3))
{
if (h1>=168 && h2>=168 && h3>=168)
printf("NO CRASH\n");
else
{
printf("CRASH ");
if (h1<168)
{
printf("%d\n",h1);
break;
}
if (h2<168)
{
printf("%d\n",h2);
break;
}
if (h3<168)
{
printf("%d\n",h3);
break;
}
}
}
return 0;
}
6 hdu 1038
#include <stdio.h>
#define P 3.1415926
int main()
{
double diameter, revolutions, time;
double sum, zong;
int n = 0;
while(~scanf("%lf%lf%lf",&diameter,&revolutions,&time))
{
if (revolutions==0) break;
n++;
sum = 0;
zong = 0;
sum = diameter / 12 / 5280 * P * revolutions * time;
zong = sum/time;
sum = zong / time * 60 * 60;
printf("Trip #%d: %.2lf %.2lf\n",n,zong,sum);
}
return 0;
}
7 hdu 1039
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <assert.h>
#define MAX_LEN 1111
char psd[MAX_LEN];
const char end[] = "end";
const char s1[] = "aeiou";
int c1(void)
{
int i,j;
for(i=0;psd[i]!='\0';i++)
for(j=0;s1[j]!='\0';j++)
if(psd[i]==s1[j])
return 1;
return 0;
}
int c2(void)
{
int i,j,cnt1,cnt2,tag;
cnt1=cnt2=0;
for(i=0;psd[i]!='\0';i++)
{
for(j=0,tag=0;s1[j]!='\0';j++)
if(psd[i]==s1[j])
{
cnt1++;
cnt2=0;
tag=1;
break;
}
if(tag==0)
{
cnt1=0;
cnt2++;
}
if(cnt1>=3 || cnt2>=3)
return 0;
}
return 1;
}
int c3(void)
{
int i;
for(i=1;psd[i]!='\0';i++)
if(psd[i]==psd[i-1])
if(psd[i]!='e' && psd[i]!='o')
return 0;
return 1;
}
int main(void)
{
#ifndef ONLINE_JUDGE
assert(freopen("1039.in","r",stdin));
#endif
while(scanf("%s",psd),strcmp(psd,end))
{
if(c1() && c2() && c3())
printf("<%s> is acceptable.\n",psd);
else
printf("<%s> is not acceptable.\n",psd);
}
return 0;
}
8 hdu 1047
大正整數加。
腦抽用string。。。越寫越醜。。。還好一遍a了，避免了重敲的宿命。。
#include <iostream>
#include <string>
#include <vector>
#include <algorithm>
#define clr(x, k) memset((x), (k), sizeof(x))
const int N = 1000;
using namespace std;
//char a[N], x[N];
string x, a;
int l;
void re(string &x)
{
int i = 0;
int j = x.length() -1;
while (i<j)
swap(x[i++], x[j--]);
}
void add(string &a, string &x)
{
int i, j, ma, mi;
ma = a.length();
mi = x.length();
if (ma<mi)
{
swap(ma, mi);
swap(a, x);
}
for (i=0;i<mi;++i) a[i] += x[i]-'0';
for (i=0;i<ma;++i)
if (a[i]>'9')
{
if (i+1<ma) a[i+1] += (a[i]-'0')/10;
else a.insert(i+1,1,(a[i]-'0')/10+'0');
a[i] -= 10;
}
}
int main()
{
int n;
scanf("%d", &n);
while (n--)
{
l = 1;
a = "0";
while (cin>>x)
{
re(x);
if (x == "0")
{
re(a);
cout<<a<<endl;
if (n!=0) cout<<endl;
break;
}
add(a, x);
}
}
return 0;
}
9 hdu 1856 ans初值爲1，wa了幾次才找到。。。
#include <iostream>
using namespace std;
const int N = 10000002;
int n, m;
int f[N], r[N];
int find(int x)
{
if (f[x]!=x) f[x] = find(f[x]);
return f[x];
}
int main()
{
int x, y, i, ans;
while (scanf("%d", &n)!=EOF)
{
for (i=0;i<=N;++i)
{
f[i] = i;
r[i] = 1;
}
ans = 1;
for (i=0;i<n;++i)
{
scanf("%d %d", &x, &y);
x = find(x);
y = find(y);
if (x>y) swap(x, y);
if (x!=y)
{
f[y] = x;
r[x] += r[y];
if (r[x]>ans) ans = r[x];
}
}
printf("%d\n", ans);
}
return 0;
}
10 hdu 1060
題目大意是輸入N，求N^N的最高位數字。1<=N<=1,000,000,000
N^N = a*10^x;
我們要求的最右邊的數字就是(int)a，即a的整數部分
兩邊同時取以10爲底的對數 lg(N^N) = lg(a*10^x) ;
N*lg(N) = lg(a) + x;
N*lg(N) - x = lg(a)
a = 10^(N*lg(N) - x);
現在就只有x是未知的了，如果能用n來表示x的話，這題就解出來了。
又因爲，x是N^N的位數-1。比如 N^N = 1200 ==> x = 3;
實際上就是 x 就是lg(N^N) 向下取整數，表示爲[lg(N^N)]
a = 10^(N*lg(N) – [lg(N^N)]);
然後(int)a 就是答案了。
#include<iostream>
#include<cstring>
#include<cmath>
using namespace std;
int main(){
int t;
__int64 ans,n;
double m;
scanf("%d",&t);
while(t--){
scanf("%I64d",&n);
m=n*log10(n+0.0);
m-=(__int64)m;
ans=pow((double)10,m);
printf("%I64d\n",ans);
}
return 0;
}11 hdu 1061 N^N%10
#include <cstdio>
#include <cstdlib>
#define I64 __int64
int Fuction(I64 n)
{
int res = 1;
I64 b = n;
if(!n) // n==0, 輸出1
return 1;
if(!(n % 10))
return 0;
while(b)
{
if(b & 1)
{
res *= n;
res %= 10; // 要取模，否則溢出
}
n *= n;
n %= 10; // 要取模，否則溢出
b >>= 1;
}
return res % 10;
}
int main()
{
int t;
I64 n;
scanf("%d", &t);
while(t--)
{
scanf("%I64d", &n);
printf("%d\n", Fuction(n));
}
return 0;
}
//題目分析：求n^n的個位數，只要根據每一個數的冪的週期性規律，就行了
#include <stdio.h>
int main()
{
int m,n,a;
int s[10][4]={{0},{1},{6,2,4,8},{1,3,9,7},{6,4},{5},{6},
{1,7,9,3},{6,8,4,2},{1,9}};
scanf("%d",&m);
while(m--)
{
scanf("%d",&n);
a=n % 10;
if(a==0||a==1||a==5||a==6)
printf("%d\n",a);
else if(a==4||a==9)
printf("%d\n",s[a][n%2]);
else if(a==2||a==3||a==7||a==8)
printf("%d\n",s[a][n%4]);
}
return 0;
}
