签到 2016.6.9
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1、CodeForces 337A Puzzles
题意：
①在有m个整数的集合中
②找到有n个整数的子集
③再比较这些子集中元素的最大值与最小值的差值
④输出最小的差值
解题思路：
①将集合中的元素从大到小排序，就可以把集合看成一个有序的序列
②比较长度为n的子序列的 “头” 和 “尾” 的差
我：
训练开始不久很多人都AC了
面对这种题我的思绪已经混乱了…
看了别人代码才懂的
用sort做的时候写比较函数写错了，写成下面这样
int cmp_num(int x, int y){ return (x - y);}WA一次
练得太少，尤其是思维锻炼
#include <iostream>#include <cstdio>#include <algorithm>using namespace std;const int maxn = 50 + 5;int f[maxn];int cmp_num(int x, int y){ return (x > y);}int main(){// freopen("in.txt", "r", stdin); int n, m; while (cin>>n>>m) { for (int i=0; i<m; ++i) { cin>>f[i]; } sort(f, f+m, cmp_num); int Min = INT_MAX; for (int i=0, j=n-1; j<m; ++i,++j) { int t = f[i] - f[j]; if (t < Min) { Min = t; } } cout<<Min<<endl; } return 0;}#include <iostream>#include <cstdio>#include <cstdlib>using namespace std;const int maxn = 50 + 5;int f[maxn];int cmp_num(const void *a, const void *b){ return *(int *)b - *(int *)a;}int main(){// freopen("in.txt", "r", stdin); int n, m; while (cin>>n>>m) { for (int i=0; i<m; ++i) { cin>>f[i]; } qsort(f, m, sizeof(f[0]), cmp_num); int Min = INT_MAX; for (int i=0, j=n-1; j<m; ++i,++j) { int t = f[i] - f[j]; if (t < Min) { Min = t; } } cout<<Min<<endl; } return 0;}2、CodeForces 520A Pangram
题意：
给一个长度为n的字符串，判断26个英文字母（不论大小写）是否都出现过
#include <iostream>#include <cstdio>#include <cmath>#include <cstring>using namespace std;char s[150];int vis[30];int main(){// freopen("in.txt", "r", stdin); int n; while (cin>>n) { memset(vis, 0, sizeof(vis)); int Count = 0; for (int i=0; i<n; ++i) { cin>>s[i]; if (s[i]>='A' && s[i]<='Z') { s[i] += ('a' - 'A'); } if (!vis[s[i]-'a']) { vis[s[i]-'a'] = 1; ++Count; } } if (Count == 26) { cout<<"YES"<<endl; } else { cout<<"NO"<<endl; } } return 0;}3、CodeForces 635A Orchestra
题意：
①给一个r行c列的图
②对图中的n个点进行标记
③找出至少包含k个被标记点的子矩形
解题思路：暴力
①遍历图中所有的点
②找到以该点为左上角的所有矩形
③再遍历这些矩形中的所有点进行判断
我：
CF的A题，搞了几个小时…..
注意有可能输入的k可能和下面定义的k混淆
#include <iostream>#include <cstring>#include <cstdio>using namespace std;const int maxn = 10 + 5;int Graph[maxn][maxn];int main(){// freopen("in.txt", "r", stdin); int r, c, n, K; while (cin>>r>>c>>n>>K) { memset(Graph, 0, sizeof(Graph)); int x, y; for (int i=0; i<n; ++i) { cin>>x>>y; Graph[x][y] = 1; } int num = 0; int i, j, k, l, a, b; for (i=1; i<=r; ++i) { for (j=1; j<=c; ++j) { for (k=i; k<=r; ++k) { for (l=j; l<=c; ++l) { int Count = 0; for (a=i; a<=k; ++a) { for (b=j; b<=l; ++b) { if (Graph[a][b] == 1) { ++Count; } } } if (Count >= K) {// cout<<i<<" "<<j<<" - "<<k<<" "<<l<<endl; ++num; } } } } } cout<<num<<endl; } return 0;}4、HDU 1597 find the nth digit
感觉挂这道题是用来练习二分的，但是我们大多人都是用了比较简单的数学方法
#include <iostream>#include <cstdio>using namespace std;int main(){// freopen("in.txt", "r", stdin); int K; while (cin>>K) { while (K--) { int N; cin>>N; int x = 1; while (N > x) { N -= x; ++x; } if (N % 9 == 0) { cout<<9<<endl; } else { cout<<N%9<<endl; } } } return 0;}5、hihoCoder 1245 王胖浩与三角形
题意：
①有一个三角形，三边长为a,b,c
②可以增加三条边的边长，增加的总长度不能超过l
③求三角形最大的面积
解题思路：
让三个边的差值尽可能的小
我：
第一次AC:
提交时间:2015-12-22 17:17:25
第二次AC:
提交时间:2016-03-29 15:27:38
#include <iostream>#include <iomanip>#include <math.h>using namespace std;double S(double a,double b,double c){ double p; double s; p = (a + b + c) / 2.0; s = sqrt(p * (p-a) * (p-b) * (p-c)); return s;}int main(){ int T; double a,b,c,l,s,M,L; double area; int i; cout.setf(ios::fixed); while (cin>>T) { while (T--) { cin>>a>>b>>c>>l; s = L = a; if (b > L) { L = b; } if (b < s) { s = b; } if (c > L) { L = c; } if (c < s) { s = c; } M = a + b + c - s - L; if (l-(M-s) > 0) { l -= (M-s); s = M; } else { s += l; l = 0; } if (l>0.0 && l>(2.0*(L-M))) { l -= (2.0*(L-M)); s = M = L; } else { s += (l/2.0); M += (l/2.0); l = 0; } if (l>0.0) { s += (l/3.0); M += (l/3.0); L += (l/3.0); }// cout<<s<<" "<<M<<" "<<L<<endl; area = S(s,M,L); cout<<setprecision(10)<<area<<endl; } } return 0;}#include <iostream>#include <cstdio>#include <cmath>#include <iomanip>using namespace std;#define Max(a, b) ((a > b) ? a : b)#define Min(a, b) ((a < b) ? a : b)double S(double a, double b ,double c);//double area(double x1, double y1, double x2, double y2, double x3, double y3);int main(){// freopen("in.txt", "r", stdin); int T; cout.setf(ios::fixed); while (cin>>T) { while (T--) { double a, b, c, l; cin>>a>>b>>c>>l; double high, mid, low; high = Max(Max(a, b), Max(b, c)); low = Min(Min(a, b), Min(b, c)); mid = (a+b+c) - (high+low); double t = mid - low; if (l > t) { l -= t; low = mid; } else { low += l; l = 0; } t = 2.0 * (high - mid); if (l > t) { l -= t; low = mid = high; } else { low += l/2.0; mid += l/2.0; l = 0; } if (l > 0) { low += (l/3.0); mid += (l/3.0); high += (l/3.0); } cout<<setprecision(10)<<S(low, mid, high)<<endl; } } return 0;}double S(double a, double b, double c){ double t; double s; t = (a + b + c) / 2.0; s = sqrt(t * (t-a) * (t-b) * (t-c)); return s;}[6、团体程序设计天梯赛-练习集 L2-008 最长对称子串](https://www.patest.cn/contests/gplt/L2-008)------------------------------------------------------------------------暴力搜include
include
include
include
using namespace std;
typedef long long ll;
const int maxn = 1000 + 10;
char s[maxn];
int main()
{
gets(s);
int Max = 1;
int len = strlen(s);
for (int i = 0; i < len; ++i) {
int t;
if (i+1 < len && s[i] == s[i+1]) {
t = 2;
for (int j = i-1, k = i+2; j>=0 && k < len && s[j] == s[k]; –j, ++k) {
t += 2;
}
if (t > Max) {
Max = t;
}
}
if (i+2 < len && s[i] == s[i+2]) {
t = 3;
for (int j = i-1, k = i+3; j>=0 && k < len && s[j] == s[k]; –j, ++k) {
t += 2;
}
if (t > Max) {
Max = t;
}
}
}
printf(“%d\n”, Max);
return 0;
}
[7、团体程序设计天梯赛-练习集 L1-006 连续因子](https://www.patest.cn/contests/gplt/L1-006)------------------------------------------------------------------------暴力搜时间复杂度O(sqrt(n))include
include
include
include
include
include
include
include
include
include
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef unsigned int uint;
const double eps = 1e-8;
const int INF = 0x7fffffff;
const int maxn = 1000;
int vis[maxn];
int num[maxn];
int main()
{
int N;
scanf(“%d”, &N);
int t = sqrt(N) + 1;
bool flag = true;
int Max = 0;
int key;
int Max_t;
for (int i = 2; i < t; ++i) {
Max_t = 0;
if (N%i == 0) {
flag = false;
int mul = i;
int add = 1;
while (N%mul == 0) {
++Max_t;
mul *= (i+add);
++add;
}
if (Max_t > Max) {
Max = Max_t;
key = i;
}
}
}
if (flag) {
printf(“1\n”);
printf(“%d\n”, N);
} else {
printf(“%d\n”, Max);
printf(“%d”, key);
for (int i = 0; i < Max - 1; ++i) {
printf(“*%d”, ++key);
}
printf(“\n”);
}
return 0;
}
[8、团体程序设计天梯赛-练习集 L1-009 N个数求和](https://www.patest.cn/contests/gplt/L1-009)------------------------------------------------------------------------注意输出细节include
include
include
include
include
include
include
include
include
include
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef unsigned int uint;
const int mod = 1e9 + 7;
const int INF = 0x7fffffff;
const int maxn = 100 + 10;
ll a[maxn], b[maxn];
ll gcd(ll a, ll b);
int main()
{
int N;
scanf(“%d”, &N);
for (int i = 0; i < N; ++i) {
scanf(“%lld/%lld”, &a[i], &b[i]);
}
ll d = abs(gcd(a[0], b[0]));
ll ans_a = a[0] / d;
ll ans_b = b[0] / d;
for (int i = 1; i < N; ++i) {
ll lcm = ans_b / gcd(ans_b, b[i]) * b[i];
ans_a = ans_a * (lcm / ans_b) + a[i] * (lcm / b[i]);
ans_b = lcm;
d = abs(gcd(ans_a, ans_b));
ans_a /= d;
ans_b /= d;
}
ll t1 = ans_a / ans_b;
ll t2 = ans_a % ans_b;
if (t1 != 0 && t2 != 0) {
printf(“%lld %lld/%lld\n”, t1, t2, ans_b);
} else if (t2 != 0) {
printf(“%lld/%lld\n”, t2, ans_b);
} else if (t1 != 0) {
printf(“%lld\n”, t1);
} else {
printf(“0\n”);
}
return 0;
}
ll gcd(ll a, ll b)
{
return (a%b == 0) ? b : gcd(b, a%b);
}
[9、CodeForces_680A Bear and Five Cards](http://acm.hust.edu.cn/vjudge/problem/viewProblem.action?id=399076)------------------------------------------------------------------------题意：输入 5 个数，求减去 2 个或 3 个相同的数后的最小值include
include
include
include
include
include
include
include
include
include
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef unsigned int uint;
const int mod = 1e9 + 7;
const int INF = 0x7fffffff;
int t[10];
int vis[110];
bool vis_t[110];
int main()
{
memset(vis, 0, sizeof(vis));
memset(vis, false, sizeof(vis_t));
int sum = 0;
for (int i = 1; i <= 5; ++i) {
scanf(“%d”, &t[i]);
++vis[t[i]];
sum += t[i];
}
int Min = sum;
for (int i = 1; i < 5; ++i) {
if (vis[t[i]] >= 3 && !vis_t[t[i]]) {
if (Min > sum - t[i] * 3) {
Min = sum - t[i] * 3;
}
vis_t[t[i]] = true;
} else if (vis[t[i]] >= 2 && !vis_t[t[i]]) {
if (Min > sum - t[i] * 2) {
Min = sum - t[i] * 2;
}
vis_t[t[i]] = true;
}
}
printf(“%d\n”, Min);
return 0;
}
[10、CodeForces_680B Bear and Finding Criminals](http://acm.hust.edu.cn/vjudge/problem/viewProblem.action?id=399032)------------------------------------------------------------------------题意：给出 n 个城市中每个城市罪犯的数量（最多为 1）和警察的位置若警察知道与他隔某个距离的位置有多少个罪犯（这个位置可能为2）求他能锁定多少个罪犯解题思路：首先他的位置若有罪犯，他就可以锁定一个然后左右两边距离为 n 的位置都有城市，那么只有两个城市都有罪犯时才能锁定两个罪犯若只有左边或右边有城市，那么只要这个城市有罪犯就可以锁定include
include
include
include
include
include
include
include
include
include
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef unsigned int uint;
const int mod = 1e9 + 7;
const int INF = 0x7fffffff;
const int maxn = 100 + 10;
int t[maxn];
int main()
{
int n, a;
scanf(“%d%d”, &n, &a);
for (int i = 1; i <= n; ++i) {
scanf(“%d”, &t[i]);
}
int Count = 0;
if (t[a] == 1) {
++Count;
}
int Left = a-1, Right = a+1;
while (Left >= 1 || Right <= n) {
if (Left >= 1 && Right <= n) {
if (t[Left] == 1 && t[Right] == 1) {
Count += 2;
}
–Left;
++Right;
} else if (Left >= 1) {
if (t[Left] == 1) {
++Count;
}
–Left;
} else if (Right <= n) {
if (t[Right] == 1) {
++Count;
}
++Right;
}
}
printf(“%d\n”, Count);
return 0;
}
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