guojiaqi007的专栏

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;

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

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,

https://www.patest.cn/contests/pat-a-practise/1007#include #include int main(){ int a[10000],n,maxSum=INT_MIN,maxBegin,maxEnd=0,sum=INT_MIN,begin = 0; scanf("%d", &n); for (int i = 0; i

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

https://www.patest.cn/contests/pat-t-practise/1015这道题开始想用bfs，但是状态转移和visit不好确定。看了http://blog.youkuaiyun.com/jtjy568805874/article/details/53610545，才知道用dp很简单。也是采用了逆向思维，想如何从 t 装变为 s，过程是把t的前半部分tHead

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[

https://www.patest.cn/contests/pat-t-practise/1017这道题首先自己写了一个O(n^2)的dp 过了。#include #include #include #include using namespace std;#define rep(i,j,k) for(int i=j;i<=k;i++)#define per(i,j

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=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=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

http://ac.jobdu.com/problem.php?pid=1209这是一个巧合装满的0-1背包问题。#include #include int main(){ int m,n; int dp[105],stamp[21]; while(scanf("%d",&m) !=EOF ){ scanf("%d",&n

https://www.patest.cn/contests/pat-a-practise/1101参考了：http://www.sigmainfy.com/blog/pat-advanced-1101-quick-sort.html用了动态规划，记录每个元素左边的最大值，右边的最小值。查看当前元素是否比左边的最大大，比右边的最小小。#include #inc

http://ac.jobdu.com/problem.php?pid=1160参考了1.http://blog.youkuaiyun.com/qq276592716/article/details/6851673#include #include int func(int m,int n){ if(m < 0) return 0; //There is n

http://ac.jobdu.com/problem.php?pid=1480#include #include #include int main(){ //freopen("in/1480.in","r",stdin); int dp[1005],a[1005],n; while(scanf("%d",&n) !=EOF){

https://www.patest.cn/contests/pat-a-practise/1117#include #include using namespace std;int a[1000001];int main(){ fill(a, a + 100001, 0); int n,len,maxLen = 0; scanf("%d",&n); for (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(){

https://www.patest.cn/contests/pat-t-practise/1002有条件的动态规划：先要排序，保证deadline靠前的task先考略，否则可能会被排在前面的deadline晚但是profit大的task淹没而不能被考虑。#include #include #include using namespace std;typede

https://www.patest.cn/contests/pat-a-practise/1042hash 一下顺序， 然后就是dp了。#include #include using namespace std;#define rep(i,j,k) for(int i=j;i<=k;i++)#define inone(i) scanf("%d",&i)const int