1.zoj-4049
简单的进程模拟,大部分情况下可以直接出答案,当进入死循环的时候,不难发现,循环中所得值会出现重复,因此可视重复为死循环的标志,使用一个bool数组进行标记即可,代码如下:
#include <iostream>
#include<cstring>
using namespace std;
const int N = 10100;
const int Mod = 256;
bool dp[N][257];
struct node {
char op[4];
int v, k;
}p[N];
bool check(int pos, int num)
{
if (dp[pos][num]) return false;
else return true;
}
int main()
{
int t;
scanf("%d", &t);
while (t--)
{
memset(p, 0, sizeof(p));
memset(dp, false, sizeof(dp));
int n;
scanf("%d", &n);
for (int i = 1; i <= n; i++)
{
scanf("%s", p[i].op);
if (strcmp(p[i].op, "add") == 0) scanf("%d", &p[i].v);
else scanf("%d%d", &p[i].v, &p[i].k);
}
int num = 0, pos = 1;
bool flag = true;
while (pos <= n && flag)
{
flag = check(pos, num);
if (!flag) continue;
dp[pos][num] = true;
if (strcmp(p[pos].op, "add") == 0)
{
num = (num + p[pos].v) % Mod;
pos++;
}
else if (strcmp(p[pos].op, "beq") == 0)
{
if (num == p[pos].v) pos = p[pos].k;
else pos++;
}
else if (strcmp(p[pos].op, "bne") == 0)
{
if (num != p[pos].v) pos = p[pos].k;
else pos++;
}
else if (strcmp(p[pos].op, "blt") == 0)
{
if (num < p[pos].v) pos = p[pos].k;
else pos++;
}
else
{
if (num > p[pos].v) pos = p[pos].k;
else pos++;
}
}
if (flag) puts("Yes");
else puts("No");
}
return 0;
}
2.zoj-4057
通过分析可得,最短的序列的二进制位一定是要相同的,这样最高位异或后皆为0,一定会比序列中最小的值还要小,所以只要求出二进制位数相同的最多的序列即可,代码如下:
#include<cstdio>
#include<cmath>
#include<iostream>
#include<string>
#include<cstring>
#include<algorithm>
using namespace std;
int a[35];
void init()
{
for (int i = 1; i <= 30; i++)
{
a[i] = (int)(pow(2, i));
}
}
int main()
{
int t;
init();
scanf_s("%d", &t);
while (t--)
{
int n;
scanf_s("%d", &n);
int sum[35] = { 0 };
for (int i = 0; i<n; i++)
{
int x;
scanf_s("%d", &x);
for (int j = 1; j <= 30; j++)
{
if (x<a[j])
{
sum[j]++;
break;
}
}
}
int ans = -1;
for (int i = 1; i <= 30; i++)
{
ans = max(ans, sum[i]);
}
printf("%d\n", ans);
}
return 0;
}
3.zoj-4056
可先画时间轴,不难发现,整个时间轴其实是由多次循环得到的,于是我们可以先求出两个时间的最小公倍数确定循环,因为每次灯泡只维持(v+0.5)s,所以将一个循环中的两个时间的倍数压进数组排序去重,灯没亮的时候要花一次去按灯,循环中的算出来后还要跑一次多出来的不在循环中的即可,代码如下:
#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>
#include<string>
#include<algorithm>
#include<vector>
using namespace std;
#define int long long
vector<long long>V;
int gcd(int a, int b)
{
if (b == 0)
return 1;
else
return gcd(b, a%b);
}
int main()
{
int T;
scanf("%lld", &T);
while (T--)
{
int a, b, c, d, v, t;
scanf("%lld%lld%lld%lld%lld%lld", &a, &b, &c, &d, &v, &t);
long long te = gcd(a, c);
long long lcm = a * c / te;
V.clear();
for (int i = 0; i <= lcm; i += a) V.push_back(i);
for (int i = 0; i <= lcm; i += c) V.push_back(i);
sort(V.begin(), V.end());
V.erase(unique(V.begin(), V.end()), V.end());
int tmp = 0;
for (int i = 1; i < V.size(); i++)
{
if (V[i] - V[i - 1] > v) tmp++;
}
long long ans = (t / a) * b + (t / c) * d + b + d - 1;
long long cur = t / lcm;
ans = ans - cur * tmp;
long long la = t % lcm;
for (int i = 1; V[i] <= la; i++)
{
if (V[i] - V[i - 1] > v) ans--;
}4
cout << ans << endl;
}
}