题目地址

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注：有的题(第三个)并没有ac,先放个错误答案。以及好多细节处理的并不好

1. F

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstring>
#include <map>using namespace std;typedef pair<int,int> PII;
int row1[6],col1[6],a1,b1; //行、列、主对角线、次对角线.
int row2[6],col2[6],a2,b2; //行、列、主对角线、次对角线.
map<int,PII> mp1,mp2;
string s[27];void solve(){memset(row1,0,sizeof row1);memset(row2,0,sizeof row2);memset(col1,0,sizeof col1);memset(col2,0,sizeof col2);a1 = 0,b1 = 0,a2 = 0,b2 = 0;string ss;for(int i=1;i<=5;i++){for(int j=1;j<=5;j++){cin >> ss;int x = (ss[0]-'0')*10+(ss[1]-'0');mp1[x] = make_pair(i,j);
//             cout << x << " ";}
//         cout << endl;}for(int i=1;i<=5;i++){for(int j=1;j<=5;j++){cin >> ss;int x = (ss[0]-'0')*10+(ss[1]-'0');mp2[x] = make_pair(i,j);
//             cout << x << " ";}
//         cout << endl;}for(int i=0;i<25;i++)    cin >> s[i];
//     for(int i=0;i<25;i++){
//         ss = s[i];
//         int x = (ss[0]-'0')*10+(ss[1]-'0');
//         cout << x << " ";
//     }bool f1 = 0,f2 = 0;for(int i=0;i<25;i++){ss = s[i];int x = (ss[0]-'0')*10+(ss[1]-'0');if(++row1[mp1[x].first]==5){f1 = 1;}if(++col1[mp1[x].second]==5){f1 = 1;}if(mp1[x].first==mp1[x].second){a1++;if(a1==5)    f1 = 1;}if(mp1[x].first+mp1[x].second==5){b1++;if(b1==5)    f1 = 1;}if(++row2[mp2[x].first]==5){f2 = 1;}if(++col2[mp2[x].second]==5){f2 = 1;}if(mp2[x].first==mp2[x].second){a2++;if(a2==5)    f2 = 1;}if(mp2[x].first+mp2[x].second==5){b2++;if(b2==5)    f2 = 1;}if(f1 || f2)    break;}if(f1==1&&f2==1)    puts("0");else if(f1==1)    puts("1");else if(f2==1)    puts("2");
}int main()
{int t;scanf("%d",&t);while(t--){solve();}return 0;
}

2. 螺旋矩阵

// 71*71#include <iostream>
#include <cstdio>
#include <cstring>using namespace std;char s[80][80];void solve(){for(int i=0;i<=80;i++)for(int j=0;j<=80;j++)s[i][j] = '.';string ss;cin >> ss;int len = ss.size();s[36][36] = ss[0];//     // 按照右上左下的顺序int i = 36,j = 36;int d = 4; // 1,2,3,4分别代表右上左下for(int k=1;k<len;k++){if(d==4){if(s[i][j+1]!='.'){ //向下走.s[++i][j] = ss[k];}else{          //向右走.d = 1;s[i][++j] = ss[k];}}else if(d==1){if(s[i-1][j]!='.'){ //向右走.s[i][++j] = ss[k];}else{          //向上走.d = 2;s[--i][j] = ss[k];}            }else if(d==2){if(s[i][j-1]!='.'){ //向上走.s[--i][j] = ss[k];}else{          //向左走.d = 3;s[i][--j] = ss[k];}            }else if(d==3){if(s[i+1][j]!='.'){ //向左走.s[i][--j] = ss[k];}else{          //向下走.d = 4;s[++i][j] = ss[k];}            }          }//    for(int i=30;i<=40;i++){
//        for(int j=30;j<=40;j++)
//            printf("%c",s[i][j]);
//        printf("\n");
//	}
//	cout << "d=" << d << endl;// 右上左下.
//  if((s[i-1][j-1]!='.'&&s[i+1][j-1]!='.') || (s[i-1][j+1]!='.'&&s[i+1][j+1]!='.')|| (s[i-1][j-1]!='.'&&s[i-1][j+1]!='.') || (s[i+1][j-1]!='.'&&s[i+1][j+1]!='.')){}else if((s[i-1][j]!='.'&&s[i-1][j+1]!='.'&&s[i][j+1]=='.')||(s[i+1][j]!='.'&&s[i+1][j+1]!='.'&&s[i][j+1]=='.')){d = 1;}else if((s[i][j-1]!='.'&&s[i-1][j-1]!='.'&&s[i-1][j]=='.')||(s[i][j+1]!='.'&&s[i-1][j+1]!='.'&&s[i-1][j]=='.')){d = 2;}else if((s[i+1][j-1]!='.'&&s[i+1][j]!='.'&&s[i][j-1]=='.')||(s[i-1][j-1]!='.'&&s[i-1][j]!='.'&&s[i][j-1]=='.')){d = 3;}else if((s[i+1][j+1]!='.'&&s[i][j+1]!='.'&&s[i+1][j]=='.')||(s[i][j-1]!='.'&&s[i+1][j-1]!='.'&&s[i+1][j]=='.')){d = 4;}if(d==1){if(s[i-1][j+1]!='.'){while(s[i-1][j+1]!='.'){s[i][++j] = ' ';}}}else if(d==2){if(s[i-1][j-1]!='.'){while(s[i-1][j-1]!='.'){s[--i][j] = ' ';}}        }else if(d==3){if(s[i+1][j-1]!='.'){while(s[i+1][j-1]!='.'){s[i][--j] = ' ';}}        }else if(d==4){if(s[i+1][j+1]!='.'){while(s[i+1][j+1]!='.'){s[++i][j] = ' ';}}   } //    for(int i=30;i<=40;i++){
//        for(int j=30;j<=40;j++)
//            printf("%c",s[i][j]);
//        printf("\n");
//	}//    for(int i=31;i<=40;i++){
//    	for(int j=32;j<=40;j++)
//            printf("%c",s[i][j]);
//    printf("\n");
//	}for(int i=1;i<=80;i++){bool f = 0;for(int j=1;j<=80;j++){if(s[i][j]!='.'){printf("%c",s[i][j]);f = 1;}}if(f)	printf("\n");}printf("\n");
}int main()
{int t;scanf("%d",&t);while(t--){solve();}return 0;
}// fjksdklsfasnkfsbfjdifsknsdfsdlg

3. 相对分子质量
   这道不知道哪里出问题了.感觉思路很清晰，写的很简洁了.

#include <bits/stdc++.h>
#include <iostream>
#include <cstdio>
#include <algorithm>
#include <map>
#include <cstring>
#include <cmath>
#include <stack>using namespace std;typedef long long ll;
map<string,ll> mp;
stack<ll> st;int main()
{ll n,m;scanf("%lld%lld",&n,&m);string s;for(ll i=1;i<=n;i++){cin >> s;ll x;scanf("%lld",&x);mp[s] = x;
//         cout << s << " " << mp[s] << endl;}for(ll j=1;j<=m;j++){string ss;cin >> ss;ll len = ss.size();s = "";for(ll i=0;i<len;i++){if(ss[i]=='('){st.push(-1);}else if(ss[i]==')'){ll sum = 0;while(st.top()!=-1){sum += st.top();st.pop();}st.pop();st.push(sum);}else if((ss[i]>='a'&&ss[i]<='z') || (ss[i]>='A'&&ss[i]<='Z')){s = s+ss[i];
//                 cout << "s=" << s << endl;if(mp[s]){
//                     cout << "mp[s]=" << mp[s] << endl;st.push(mp[s]);s = "";}}else if(ss[i]>='0'&&ss[i]<='9'){string sss = "";while(ss[i]>='0'&&ss[i]<='9'){sss += ss[i];i++;}
//                 cout << "sss=" << sss << endl;ll cnt = 0;ll mid = 1;for(ll k=sss.size()-1;k>=0;k--){cnt += mid*(sss[k]-'0');mid *= 10;}
//                 cout << "cnt=" << cnt << endl;ll sum = st.top();st.pop();sum *= (cnt);
//                 cout << "sum=" << sum << endl;st.push(sum);i--;}}ll ans = 0;while(!st.empty()){if(st.top()==-1){st.pop();continue;}ans += st.top();st.pop();}printf("%lld\n",ans);}return 0;
}/*
4 1
Ba 66
O 16
H 1
C 12
Ba((OH)2(CO3)2)3debug:(OH)10
*/

4. 统计个数

#include <iostream>
#include <algorithm>
#include <cstring>using namespace std;const int N = 210;int t, n, m;
bool g[N][N];int gcd(int a, int b){return b ? gcd(b, a % b) : a;
}int in[N];
int a[N];
int main()
{cin >> t;a[1] = 1; a[0] = 1;for (int i = 2; i <= 200; i++){a[i] = a[i - 1] * i;}while (t--){memset(g, false, sizeof g);memset(in, 0, sizeof in);scanf("%d%d", &n, &m);for (int i = 0; i < m; i++){int x, y;scanf("%d%d", &x, &y);g[x][y] = true;g[y][x] = true;in[x]++;in[y]++;}int p = 0, q = 0;for (int i = 1; i <= n; i++)for (int j = i+1; j <= n; j++){for (int k = j+1; k <= n; k++){if (g[j][i] && g[i][k]){q++;if (g[j][k]) p++;}}}int ans = 0;
// 		cout << p*3 << ' ' << q << endl;// 统计入度数.for (int i = 1; i <= n; i++){ans += (in[i] * (in[i] - 1)) / 2;}if (!p) puts("0/1");else{p = p*3, q;int q = gcd(p, ans);printf("%d/%d\n", p/q , ans/q);}}return 0;
}

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PS:模拟能力有待提升.要注重细节处理，全盘考虑想清楚.