HNUOJ13302和13308贪心枚举有俩种情况的想办法直接枚举俩种情况每次枚举比较值

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13302：

#include<algorithm>#include<iostream>#include<limits.h>#include<stdlib.h>#include<string.h>#include<complex>#include<cstring>#include<iomanip>#include<stdio.h>#include<bitset>#include<cctype>#include<math.h>#include<string>#include<time.h>#include<vector>#include<cmath>#include<queue>#include<stack>#include<list>#include<map>#include<set>#define LL long longusing namespace std;const LL mod = 1e9 + 7;const double PI = acos(-1.0);const double E = exp(1.0);const int M = 1e6 + 5;string s;int dis[M];int main(){    LL a, b;    while( cin >> a >> b ){        cin >> s;        b = a - b;        int len = s.size();        int sum = 0;        memset(dis, 0, sizeof(dis));        for(int i = 0; i < len; ++i){            if(s[i] == 'B'){                sum++;                dis[sum] = i + 1;            }        }        if(sum == 0){            puts("0");            continue;        }        LL mn = 1e11;        LL ans;        for(int i = 0; i <= sum; ++i){            ans = (LL)(i * a);            for(int j = 1; j <= sum - i; j++){                if(dis[j] > j + i)                    ans += (LL)(b * (dis[j] - j - i));            }            mn = min(mn, ans);        }        cout << mn << endl;    }    return 0;}

13308：

#include<algorithm>#include<iostream>#include<limits.h>#include<stdlib.h>#include<string.h>#include<complex>#include<cstring>#include<iomanip>#include<stdio.h>#include<bitset>#include<cctype>#include<math.h>#include<string>#include<time.h>#include<vector>#include<cmath>#include<queue>#include<stack>#include<list>#include<map>#include<set>#define LL long longusing namespace std;const LL mod = 1e9 + 7;const double PI = acos(-1.0);const double E = exp(1.0);const int M = 1e3 + 5;int a[M];int main(){    int n;    while( cin >> n ){        for(int i = 0; i < n; ++i)            cin >> a[i];        sort(a, a + n);        int mn = 1e9;        int ans;        for(int i = 0; i <= n / 2; ++i){            ans = 0;            for(int j = 0; j < i; ++j){                ans += 24 - a[n - j - 1] + a[j];            }            for(int j = i; j < n - i; j += 2){                ans += a[j + 1] - a[j];            }            mn = min(mn, ans);        }        cout << mn << endl;    }    return 0;}

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