2017年浙江工业大学大学生程序设计迎新赛预赛

挺不错的一套题 写了几道鸽了没写完 有空继续补

比赛链接 https://www.nowcoder.com/acm/contest/52#question

A.

尼姆博弈的想法 很久没写题了 博弈都忘完了 看了提示 其实很简单，，，

#include <bits/stdc++.h>
using namespace std;
#define LL long long
const int MAXN = 1e5+10;
int main()
{
ios::sync_with_stdio(false);
int T; cin >>T;
while (T--) {
int n, k, x, y; cin >> n >> k;
int res = 0;
for(int i = 1; i <= n; i++) {
cin >> x;
if(k == i) {
y = x; continue;
}
res ^= x;
}
if(res < y) cout << "Yes\n";
else cout << "No\n";
}
return 0;
}

B. 多项式求值

感觉这个题好zz啊 就是个简单的快速幂 非要 用字符串读入 而且是多组输入 debug到崩溃QAQ

#include <bits/stdc++.h>
using namespace std;

#define LL long long
const LL mod = 1e9+7;
const LL MAXN = 1e6+100;
char str[MAXN];
LL len;
LL pwo_mod(LL a, LL b)
{
LL ans = 1; a %= mod;
while(b) {
if(b&1) ans = (ans*a)%mod;
a = (a*a)%mod;
b >>= 1;
}
return ans;
}
LL get_num(LL l, LL r)
{
LL num = 0, res = 1;
for(LL i = r; i >= l; i--) {
nu
4000
m += res*(str[i]-'0');
res *= 10;
}
return num;
}
LL calcu(LL a, LL b, LL x)
{
LL res = 1ll*a*pwo_mod(x,b)%mod;
return res;
}
bool is_num(LL x)
{
if(str[x]>='0' && str[x]<='9') return true;
return false;
}
void solve(LL x)
{
LL ans = 0;
LL a, b, l, r;
LL clr;
for(LL i = 0; i < len; i++) {
if(str[i] == '-') clr = -1;
else clr = 1;
if(str[i]=='+' || str[i]=='-') i++;
l = r = i;
if(str[l] == 'x') {
a = 1, b = 1; l--, r--;
}
else {
while(r+1<len && is_num(r+1)) r++;
a = get_num(l,r); b = 1;
}
if(str[r+1] != 'x') {
b = 0; i = r;
}
else if(str[r+2] == '^') {
l = r+3, r = l;
while(r+1<len && is_num(r+1)) r++;
b = get_num(l,r); i = r;
}
else i = r+1;
ans = (ans+clr*calcu(a,b,x))%mod;
}
ans = (ans%mod+mod)%mod;
printf("%lld\n",ans);
}
int main()
{
while(~scanf("%s",str)) {
LL x, m;
len = strlen(str);
scanf("%lld",&m);
while(m--) {
scanf("%lld",&x); solve(x);
}
}

return 0;
}

C. 数的变换

数学相关 有空在补

这里写代码片

D. 简单的数据结构

用deque写下就好 list超时mmp

#include <bits/stdc++.h>
using namespace std;
#define LL long long
const int MAXN = 1e5+10;
int main()
{
ios::sync_with_stdio(false);
int n, m, x, y;
deque<int> d;
cin >> n >> m;
while(m--) {
cin >> x;
if(x == 1) {
cin >> y; d.push_front(y);
}
if(x == 2) {
d.pop_front();
}
if(x == 3) {
cin >> y; d.push_back(y);
}
if(x == 4) {
d.pop_back();
}
if(x == 5) {
reverse(d.begin(),d.end());
}
if(x == 6) {
cout << d.size() << endl;
for(int i = 0; i < d.size()-1; i++) cout << d[i] << ' ';
cout << d[d.size()-1] << "\n";
}
if(x == 7) {
sort(d.begin(),d.end());
}
}

return 0;
}