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http://tyvj.cn/p/1003模拟即可，上坡路段就加上下坡路段的时间，平路路段要*2，下坡路段就加上上坡路段的时间，时间不够了一定立即返回，否则会有问题#include#include#includeusing namespace std;int main () { int m,t,u,f,d; scanf("%d%d%d%d%d%*c", &m, &t, &u

https://www.luogu.org/problem/show?pid=1146由样例可得，翻转第i次时，第i个硬币不翻，显然最小次数即为n#include#include#includeusing namespace std;#define ms(i,j) memset(i, j, sizeof i);int n;int f[105];int main(){

https://www.luogu.org/problem/show?pid=2689由题意显然只用朝两个方向走，确定下来这两个方向以后，比对每个时刻的风向即可#include#include#include#includeusing namespace std;#define ms(i,j) memset(i, j, sizeof i);int x1,x2,y1,y2,T;

https://www.luogu.org/problem/show?pid=1847类似铺地毯，把所有攻击范围存起来再根据询问统计模拟即可#include #include #include using namespace std; #define ms(i,j) memset(i, j, sizeof i); struct node{ int x1;

3^0，3^1，3^0+3^1，3^2，3^0+3^2，3^1+3^2，3^0+3^1+3^2把i转化为2进制，i为第几项数对应于上面1 10 11 100 101 110 111找到规律，把n转为2进制即可解决#include#include#include#include#define ms(i,j) memset(i,j, sizeof i);using n

http://www.luogu.org/problem/show?pid=1328直接模拟，弄一个邻接矩阵#include#include#include#include#define ms(i,j) memset(i,j, sizeof i);using namespace std;int G[10][10] = { { 0,-1, 1, 1,-1}, { 1, 0

http://www.luogu.org/problem/show?pid=1203按题意模拟即可#include#include#include#include#define ms(i,j) memset(i,j, sizeof i);using namespace std;char ch[1000];int main(){ int n; scanf("%d\n",

http://www.luogu.org/problem/show?pid=1059这里用的桶排序，自动完成排序+去重，因为生成的数不大于1000，数据不大#include#include#include#include#define ms(i,j) memset(i, j, sizeof(i));using namespace std;int a[105];bool v[1

http://www.luogu.org/problem/show?pid=1309#sub第一次快排，之后用a,b记录输赢的集合，赢了的和输了的的相对排序一致，用归并排序#include#include#include#include#define ms(i,j) memset(i,j, sizeof i);using namespace std;struct ps{ i

http://www.luogu.org/problem/show?pid=1086由于必须采摘当前最大植株，所以这道题水了很多，当如果摘了当前点返回不够时间的话，那这个点就不能摘。#include#include#include#include#define ms(i,j) memset(i,j, sizeof i);using namespace std;struct fr

http://www.luogu.org/problem/show?pid=1181水题不解释#include#include#include#include#define ms(i,j) memset(i,j, sizeof i);using namespace std;int main(){ int n,m; int ans = 1; int to

http://www.luogu.org/problem/show?pid=1781排序即可，注意要用字符串，注意字符串的交换方法以及比较大小，n#include#include#include#include#define ms(i,j) memset(i, j, sizeof(i));using namespace std;int xh[25];char ch[25][1

https://vijos.org/p/1975处理二维数组即可，注意坐标偏移即可#include

http://www.luogu.org/problem/show?pid=1003不要用搜索，范围不行。用扫描的形式求解即可#include #include #include #define ms(i,j) memset(i, j, sizeof(i));using namespace std;int n;int x[10005], y[10005], g[10005]

http://www.luogu.org/problem/show?pid=1067看似简单的模拟题，但是许多细节要考虑。具体看代码注释，这样写比较清晰#include#include#include#include#define ms(i,j) memset(i, j, sizeof(i));using namespace std;int n;int main(){ s

http://tyvj.cn/p/1006一道简单的水题，注意X即可#include#include#includeusing namespace std;char ch[80];int main () { scanf("%s", ch); int k = 0; int tot = 0; for (int i=0;i<11;i++) { if (ch[i]>='

http://tyvj.cn/p/2772此题由于我不喜欢用栈写表达式（分明是不会），所以用个链表乱搞AC了。。#include#include#includeusing namespace std;int sz[100005];//数字 int bh[100005];//标号，链表储存 int s_n = 0;char fh[100005];//符号 int f_n =

https://www.luogu.org/problem/show?pid=1184用个map搞定。注意要getline#include#include#include#include#include#includeusing namespace std;#define ms(i,j) memset(i, j, sizeof i);int n,m;map s;stri