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2015 Fall HIT Weekly Training 2

2015-10-24 21:04 459 查看
在错过了第一次之后,终于准时参加了第二次。

18:30-20:30的新手练习题,我居然真的就做了两个小时,差点没AC

A.YY, and YY again

  读取string时会停在空格,所以用getline就行。

#include<cstdio>
#include<cstring>
#include<iostream>
using namespace std;

int main(void)
{
string st;
int x;
while(getline(cin, st)){
x = 0;
for(int i = 0; i < st.length(); ++i)
if(st[i] >= 'A' && st[i] <= 'Z')
x += st[i] - 'A' + 1;
if(x > 100) cout<<"INVALID"<<endl;
else cout<<x<<endl;

}
return 0;
}


B.Box of Bricks

  这年头,思路比这再简单的数学题真是不多见了。

计算∑(a[i]<p)(p-a[i])就行了。p是最后的高度。

#include<cstdio>

int a[100];

int main(void)
{
int N, k = 1;
int p;
int ans;
scanf("%d", &N);
while(N > 0){
if(k > 1) printf("\n");
p = 0;
ans = 0;
for(int i = 0; i < N; ++i){
scanf("%d", &a[i]);
p += a[i];
}
p /= N;
for(int i = 0; i < N; ++i)
if(a[i] < p)
ans += p - a[i];
printf("Set #%d\n", k++);
printf("The minimum number of moves is %d.\n", ans);
scanf("%d", &N);
}
return 0;
}


C.Not A Water Problem

  首先是那个诡异Hint:That isn't a real water problem,but I promise that there's only one trick.

  一看就知道这道题暗藏杀机,开始时候使用while(scanf()),妥妥的TLE。

  奥,原来就trick就是EOF啊,果断改,试了!EOF、getchar() != EOF 、(ch = getchar()) != EOF。果断都不管用,花式WA。。。

  然后恍然大悟,signed int,果然是个暗示啊,abs(-2^31)不能用int,然后换LL,还是各种WA。。。

  在最后的两分钟,开始各种尝试,终于当我换回了while(cin)的时候,它红了。

  。。。。。。

#include<cstdio>
#include<iostream>
using namespace std;

#define abs(a) ((a)>0)?(a):(-(a))

int main(void)
{
long long x;
while(cin>>x){
cout<<(abs(x))<<endl;
}
return 0;
}


D.整数的排序

  果断的sort

#include<cstdio>
#include<algorithm>
#include<iostream>
#include<cstdlib>

int a[20];

int main(void)
{
int N;
scanf("%d", &N);
for(int i = 0; i < N; ++i){
for(int j = 0; j < 10; ++j) scanf("%d", &a[j]);
std::sort(a, a+10);
for(int j = 0; j < 9; ++j) printf("%d ", a[j]);
printf("%d\n", a[9]);
}
return 0;
}


E.Fibonacci Numbers

  看见这题瞬间就想起了那年哪月,翔神教我的矩阵快速幂求斐波那契。。。REBMYSMLY

  偷偷的百度了lrj的bign,ctrl+c, ctrl+p。

  想着有lrj庇佑,这题肯定一次AC,然而。。。

  经典坑爹,教科书式的坑爹,测试数据有多组,但他就是不告诉你。

#include<cstdio>
#include<iostream>
#include<cstring>
using namespace std;

#define MAXN 3010

struct bign{
int len, s[MAXN];

bign() {
memset(s, 0, sizeof(s));
len = 1;
}

bign(int num) {
*this = num;
}

bign(const char* num) {
*this = num;
}

bign operator = (int num) {
char s[MAXN];
sprintf(s, "%d", num);
*this = s;
return *this;
}

bign operator = (const char* num) {
len = strlen(num);
for(int i = 0; i < len; i++) s[i] = num[len-i-1] - '0';
return *this;
}

string str() const {
string res = "";
for(int i = 0; i < len; i++) res = (char)(s[i] + '0') + res;
if(res == "") res = "0";
return res;
}

bign operator + (const bign& b) const{
bign c;
c.len = 0;
for(int i = 0, g = 0; g || i < max(len, b.len); i++) {
int x = g;
if(i < len) x += s[i];
if(i < b.len) x += b.s[i];
c.s[c.len++] = x % 10;
g = x / 10;
}
return c;
}

void clean() {
while(len > 1 && !s[len-1]) len--;
}

bign operator * (const bign& b) {
bign c; c.len = len + b.len;
for(int i = 0; i < len; i++)
for(int j = 0; j < b.len; j++)
c.s[i+j] += s[i] * b.s[j];
for(int i = 0; i < c.len-1; i++){
c.s[i+1] += c.s[i] / 10;
c.s[i] %= 10;
}
c.clean();
return c;
}

bign operator - (const bign& b) {
bign c; c.len = 0;
for(int i = 0, g = 0; i < len; i++) {
int x = s[i] - g;
if(i < b.len) x -= b.s[i];
if(x >= 0) g = 0;
else {
g = 1;
x += 10;
}
c.s[c.len++] = x;
}
c.clean();
return c;
}

bool operator < (const bign& b) const{
if(len != b.len) return len < b.len;
for(int i = len-1; i >= 0; i--)
if(s[i] != b.s[i]) return s[i] < b.s[i];
return false;
}

bool operator > (const bign& b) const{
return b < *this;
}

bool operator <= (const bign& b) {
return !(b > *this);
}

bool operator == (const bign& b) {
return !(b < *this) && !(*this < b);
}

bign operator += (const bign& b) {
*this = *this + b;
return *this;
}
};

istream& operator >> (istream &in, bign& x) {
string s;
in >> s;
x = s.c_str();
return in;
}

ostream& operator << (ostream &out, const bign& x) {
out << x.str();
return out;
}

int main(void)
{
int N;
bign a1, a2;
while(cin>>N){
a1 = 1;
a2 = 1;
for(int i = 3; i <= N; i += 2){
a1 = a1 + a2;
a2 = a1 + a2;
}
if(N & 1) cout<<a1<<endl;
else cout<<a2<<endl;
}
return 0;
}
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