天行健，君子以自强不息

注意当结果为0的情况#include#includeusing namespace std;int main(){    int a,b;    while(cin>>a>>b)    {        int c=fabs(a+b);    int cnt=0;    int aa=1;    char str[10];    while(c){

#define LOCAL#include #include #include #include #include using namespace std;typedef long long ll;ll gcd(ll a,ll b){ return b==0?a:gcd(b,a%b);}struct Fraction{ ll up,down;};Fract

#define LOCAL#include #include #include #include #include using namespace std;typedef long long ll;int a[1000100];bool isprime(int x){ for(int i=2;i*i<=x;i++){ if(x%i==0) return

本题关键：将字母映射成数字表示，之后用辅助数组即可完成操作#include#include#include #includeusing namespace std;const int N=54;char mp[5]={'S','H','C','D','J'};int start[N+1],rear[N+1],pos[N+1];int main(){ int k;

#define LOCAL#include #include #include #include #include using namespace std;typedef long long ll;bool isprime(int x){ for(int i=2;i*i<=x;i++){ if(x%i==0) return false; }

#define LOCAL#include #include #include #include #include #include using namespace std;typedef long long ll;bool isprime(int x){ for(int i=2;i*i<=x;i++){ if(x%i==0) return false

#define LOCAL#include #include #include #include #include #include using namespace std;typedef long long ll;bool isprime(int x){ for(int i=2;i*i<=x;i++){ if(x%i==0) return false

关键：需要输入时进行预处理，否则会超时#include#include#include #includeusing namespace std;const int MAXN=100005;int dis[MAXN],a[MAXN];int main(){ int n; cin>>n; int sum=0; for(int i=1;i<=n;i

#include#include#include #includeusing namespace std;int a[100001];int main(){ int t; cin>>t; int cnt=1; while(t--){ long long a,b,c; scanf("%lld%lld%lld",&a,&

#define LOCAL#include #include #include #include #include #include using namespace std;typedef long long ll;struct bign{ int d[1010]; int len; bign(){ memset(d,0,sizeof(d));

注意使用高精度#define LOCAL#include #include #include #include #include #include using namespace std;typedef long long ll;struct bign{ int d[1010]; int len; bign(){ memset(d,0,si

#define LOCAL#include #include #include #include #include #include #include using namespace std;const int N = 40010;const int M=26*26*26*10+1;vector selectCourse[M];typedef long long ll;

#include#include#include #includeusing namespace std;float a[1001],b[1001],c[1001];int main(){ int k1,k2; cin>>k1; int cnt=0; for(int i=0;i<k1;i++){ int ep; floa

#include#include#include #includeusing namespace std;struct Poly{ int exp; float cof;}poly[1001];float ans[2001];int main(){ int k; cin>>k; for(int i=0;i<k;i++){

#define LOCAL#include #include #include #include #include #include #include #include using namespace std;const int N=51;set st[N];int main(){ #ifdef LOCAL freopen("data.in",

#include#include#include #includeusing namespace std;int main(){ double ans=1.0; char S[3]={'W','T','L'}; char pos[4]; //输入时直接处理 for(int i=0;i<3;i++){ double temp

考查点：DFS递归，Dijkstra求最短路径思路：这道题不能直接用Dijkstra来更新need和remain必须用DFS遍历所有最短路径才能更新，写Dijkstra记得判断vis[u]==false，路径用pre记录前驱节点，因为有多条最短路径，必须用Vector数组来保存，Dijkstra主要涉及的就是pre的更新，DFS每次递归到边界0处时候遍历temp路径计算need和remain，

#include #include #include using namespace std;int main(){ int n; cin>>n; char e[16],l[16]; int hmin=24,mmin=60,smin=60; int hmax=0,mmax=0,smax=0; while(n--){

#include #include #include using namespace std;struct person{ char id[15]; char name[15]; int grade; char sex;}high,low,temp;void init(){ high.grade=-1; low.grade=101;

#include#include#include #include#include using namespace std;char str[82];int main(){ scanf("%s",str); int n=strlen(str); int ans=0,k=3; int n2; while(k<n){ if(3

#define LOCAL#include #include #include #include #include #include #include #include using namespace std;int n;string deal(string s,int& e){ int k=0; while(s.length()>0&&s[0]=='0')

#define LOCAL#include #include #include #include #include #include #include #include #include using namespace std;string unitDigit[13]={ "tret", "jan", "feb", "mar", "apr", "may", "jun

注意边界数据0的输出#include #include #include using namespace std;int main(){ int n,b; cin>>n>>b; int a[40]; int k=0; int ans=n; while(n){ a[k++]=n%b; n=n/b;

进制转化，通过开一个字符数组表示对应关系简化判断#include #include #include using namespace std;int main(){ char radix[13]={'0','1','2','3','4','5','6','7','8','9','A','B','C'}; int a,b,c; cin>>a>>b>>c;

注意溢出，这里出现是计算结果产生溢出#include #include #include using namespace std;int main(){ long long g,s,k; scanf("%lld.%lld.%lld",&g,&s,&k); long long cost=17*29*g+29*s+k; scanf("%lld.%lld.

#define LOCAL#include #include #include #include #include #include #include #include #include using namespace std;int main(){ #ifdef LOCAL freopen("data.in","r",stdin);

#define LOCAL#include #include #include #include #include #include #include #include #include using namespace std;bool check(char c){ if(c>='A'&&c<='Z') return true; if(c>='0'&&c<=

关键：map和set结合使用#define LOCAL#include #include #include #include #include #include #include #include #include using namespace std;map > mpTitle,mpAuthor,mpKey,mpPub,mpYear;void query(map

#define LOCAL#include #include #include #include #include #include #include #include #include using namespace std;int st[10010];int a[1010];int main(){ #ifdef LOCAL freopen(

#include #include #include #include using namespace std;int main(){ char week[7][5]={"MON","TUE","WED","THU","FRI","SAT","SUN"}; int ans=0; char str1[70],str2[70],str3[70],str4[70];

#include #include #include #include using namespace std;int main(){ char str[10010]; gets(str); char ans[10010]; int len=strlen(str); if(str[0]=='-') cout<<'-'; ans[0]=s

#define LOCAL#include #include #include #include #include #include #include #include #include #include using namespace std;struct mouse{ int w; int r;}mouse[1010];int main(){

#include #include #include #include using namespace std;char change[10][10]={ "zero","one","two","three","four","five","six", "seven","eight","nine"};int main(){ char str[110];

#include #include #include #include using namespace std;struct note{ char usr[11]; char pwd[11];}mem[1001];int flag[1001];int main(){ int n; cin>>n; int cnt=0; int f=

关键：旋转字符串，转化为最长前缀#include #include #include #include using namespace std;int main(){ int n; cin>>n; char str[101][260]; int lenmin=260; getchar(); for(int i=0;i<n;i++)

关键：字符串问题，注意边界0的处理，各种特殊情况#include #include #include #include using namespace std;int main(){ char change[11][10]={"ling","yi","er","san","si","wu","liu","qi","ba","jiu" }; char wei[5]

考查点：Dijkstra#include #include #include #include #include #include #include #include #include #include #include #define FOR(i, x, y) for(int i = x; i <= y; i++)#define rFOR(i, x, y) for(

考查点：Dijkstra算法思路：第一次输入少了&，第二次本题是1到n+m，缺少了+m，还有id转化两次写错，第一次==写错成=，第二次忽视了数字为多位数的情况，浪费时间最多的是调用ds时应该从起点s+n而不是s+m，另外，本题最求最大的最近距离，变量应设为里dmax里面才是dmin#define LOCAL#include #include #include #include

#include #include #include #include using namespace std;struct Student{ char id[15]; int score; int local_num; int local_rank;}stu[30010];bool cmp(Student a,Student b){ if(

考查点：Dijkstra提交情况：输入处理没做好，结果调了半天。。。虽然没报错，是因为输入是字符串，所以最后接受的是空串，注意这里输入是n-1而不是n。。还有以后调试多输出中间结果，比较快。本题思路直接套dj算法，在维护几个数组增加判断情况即可#define LOCAL#include #include #include #include #include #include