guojiaqi007的专栏

http://ac.jobdu.com/problem.php?pid=1548code from http://www.lai18.com/content/477808.html#include #include #include using namespace std;#define ms(x,y) memset(x,y,sizeof(x))#define i

http://ac.jobdu.com/problem.php?pid=1140#include #include static int map[8][8];static int pos[8];static int ans[92],cn;void deploy(int i, int j, int v){ int _i=i,_j=j; while(_i<8) m

http://ac.jobdu.com/problem.php?pid=1131两次动态规划 最长子序列#include #include #include int main(){ //freopen("in/1131.in","r", stdin); int height[102],n; int dpl[101],dph[101],i

http://ac.jobdu.com/problem.php?pid=1130参考了 http://www.cnblogs.com/jasonJie/p/5740646.html#include #include #include typedef struct date{ int year,month,day; friend bool oper

http://ac.jobdu.com/problem.php?pid=1119#include #include #include int main(){ //freopen("in/1119.in","r",stdin); int res[200],num[102],rlen,len,c,x; char str[102]; memset

http://ac.jobdu.com/problem.php?pid=1147好像 九度的oj judge 有问题，不对的code也可以过。建议使用 UVa的测试https://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&problem=512自己谢了bfs

http://ac.jobdu.com/problem.php?pid=1149#include #include #include typedef struct item{ char s[102]; int times; bool friend operator < (struct item a, struct item b){

http://ac.jobdu.com/problem.php?pid=1161#include static int N,Q; static char Map[3001][3001];static char Template[6][6];int myPow(int n,int q){ int m = 1; while(q--){ m*=n;

参考了 http://www.acmerblog.com/combinations-of-r-elements-6059.html#include #include void combine(int arr[], int data[], int start, int end,int idx,int r){ // has found r elements if(

http://ac.jobdu.com/problem.php?pid=1159#include #include typedef struct ant{ int pos,dir; bool friend operator < (struct ant a, struct ant b){ return a.pos < b.pos; }

http://ac.jobdu.com/problem.php?pid=1154最小生成树：堆+并查集#include #include typedef struct edge{ int l,r,w; bool friend operator < (struct edge a, struct edge b){ return a.w > b.w;

http://ac.jobdu.com/problem.php?pid=1040两种方法：#include #include #include #define MAXN 2000000static int primes[10001],nPrimes;static bool mark[MAXN];void init0(){ memset(mark,true,

http://ac.jobdu.com/problem.php?pid=1457#include #include #include #include using namespace std;bool visit[101][101][101];int n, m, s;typedef struct status { int a,b,c; int steps; void

http://ac.jobdu.com/problem.php?pid=1142此题利用了中国剩余定理参考了1.http://www.cnblogs.com/My-Sunshine/p/4830388.html2.http://blog.youkuaiyun.com/sinat_30126425/article/details/52154948代码来自于:参考2

http://ac.jobdu.com/problem.php?pid=1112#include #include #include int main(){ //freopen("in/1112.in","r",stdin); int a[26], dp[26],n; while(scanf("%d",&n)!=EOF){ for

http://ac.jobdu.com/problem.php?pid=1114#include int main(){ //freopen("in/1114.in","r",stdin); int n,dp[41],a[41]; while(scanf("%d",&n) !=EOF){ for (int i = 1; i <=n;

http://ac.jobdu.com/problem.php?pid=1072#include const int n = 15;const int v = 8 * 5 + 10 * 4 + 18 * 6;const int a[n+1] = { 0,8,8,8,8,8,10,10,10,10,18,18,18,18,18,18 };#define rep(i,j,k) fo

参考了1.王道的机试复试书#include #include #include static int p[][7] = { /* $ + - * % ( )*/ /*$ */ { 0,-1,-1,-1,-1,-1,-2}, /*+*/ { 1, 1, 1,-1,-1,-1, 1}, /*-*/ { 1, 1,

http://ac.jobdu.com/problem.php?pid=1037参考了1.王道机试指南#include #include #include #include using namespace std;static const int maxn = 200;static const int ten[4] = {1,10,100,1000}; t

http://ac.jobdu.com/problem.php?pid=1549参考了http://www.voidcn.com/blog/newchan/article/p-4594278.html贪心，用大面值肯定比用小面值 张数少#include #include #include using namespace std;#define ms(x,y) memset(x,

http://ac.jobdu.com/problem.php?pid=1550默认给最少“1”， 如果a[i] > a[i-1], 比前面多给一个。如果 a[i] 参考了http://blog.youkuaiyun.com/xiexingshishu/article/details/20319825#include #include using namespace std;#def

http://ac.jobdu.com/problem.php?pid=1552动归，分最后一个是男还是女讨论。注意当总共有0人以男结尾的排列是1种情况。 #include #define inone(i) scanf("%d",&i)const int maxn = 1e3 + 10;int m[maxn], f[maxn], n;void init(){ f[

http://ac.jobdu.com/problem.php?pid=1554copy from http://blog.youkuaiyun.com/u013491262/article/details/21406351把sum[1~i] = sum[i],  sum[i] 是以a[i]为最后一个元素的前缀和，  Map以前缀和为key， 具有相同的前缀和的idx的vector 为va

http://ac.jobdu.com/problem.php?pid=1555这道题觉得还可以再优化，暂时先水过了。#include #include #include using namespace std;#define rep(i,j,k) for(int i=j;i<=k;i++)string s;int cal(string s){ set uniq;

http://ac.jobdu.com/problem.php?pid=1171#include #include using namespace std;#define rep(i,j,k) for(int i=j;i<=k;i++)#define inone(i) scanf("%d",&i)#define infour(a,b,c,d) scanf("%d%d%d%d"

http://ac.jobdu.com/problem.php?pid=1010#include #include #include #include #include using namespace std;map Map;int get(string s){ string a; istringstream is(s); int res = 0; while

http://ac.jobdu.com/problem.php?pid=1009#include #include #include using namespace std;#define rep(i,j,k) for(int i=j;i<=k;i++)#define inone(i) scanf("%d",&i)//typedef struct nd { int v;

http://ac.jobdu.com/problem.php?pid=1008这道题告诉我们 INT_MAX, 慎用，dijkstra 有时加溢出了，你都不知道#include //#include #include using namespace std;#define rep(i,j,k) for(int i=j;i<=k;i++)#define inone(i)

http://ac.jobdu.com/problem.php?pid=1006#include #include #include using namespace std;#define rep(i,j,k) for(int i=j;i<=k;i++)bool ac(string s){ //cout << s << endl; int len = s.size()

http://ac.jobdu.com/problem.php?pid=1073#include #include #define inone(i) scanf("%d",&i)#define rep(i,j,k) for(int i=j;i<=k;i++)const int maxn = 1e6;int n;int dp[2][maxn];int get(int i,

http://ac.jobdu.com/problem.php?pid=1084照抄了 http://blog.youkuaiyun.com/jdplus/article/details/18839343参考了 http://www.cnblogs.com/yinger/archive/2012/08/17/2644312.html#include int main(){

http://ac.jobdu.com/problem.php?pid=1043有两种做法：一公式法：如code1。编码简单，但是在九度上不能ac。二比较法：如code2，比较今天和目标时间的差，推算目标时间的星期。code1:#include #include static const char* month [] { "","January", "

http://ac.jobdu.com/problem.php?pid=1111这一题我用了 stl#include #include using namespace std;int main(){ //freopen("in/1111.in","r",stdin); string str,newString, a,b; string::si

http://ac.jobdu.com/problem.php?pid=1081参考了（照抄了）http://blog.youkuaiyun.com/zxasqwedc/article/details/8763967 #include #include #include #define LL long longstatic LL a0,a1,p,q,k; typedef

http://ac.jobdu.com/problem.php?pid=1042#include #include #include int main(){ //freopen("in/1042.in","r",stdin); int dp[101][101],m,n; char a[101],b[101]; while(scanf("%s

#include #include #include typedef struct gas_station{ double price; int dis; friend bool operator < (struct gas_station a, struct gas_station b){ return a.dis < b.dis;

http://ac.jobdu.com/problem.php?pid=1085refs: 1.http://blog.youkuaiyun.com/JDPlus/article/details/18840577?locationNum=1&fps=12.http://blog.sina.com.cn/s/blog_8619a25801010wcy.htmlCode from

http://ac.jobdu.com/problem.php?pid=1087照抄了 http://blog.youkuaiyun.com/jdplus/article/details/18353667#include #include int main(){ //freopen("in/1087.in","r",stdin); int n; int

http://ac.jobdu.com/problem.php?pid=1096计算每个日期的总日子，然后相减。#include static int days = 365;static int month[13] = { 0,31,28,31,30,31,30,31,31,30,31,30,31 }; int leap(int year){ if(!

http://ac.jobdu.com/problem.php?pid=1086#include #include #include int main(){ //freopen("in/1086.in","r",stdin); long long l1,l2,l3,c1,c2,c3; int A,B; int n=0; int* a;