CF

A.Binary Protocol

题目链接

题意: 将每段连续1的个数输出，以0结尾，注意最后如果有0还要多输出一个0

#include<cstdio>#include<cmath>#include<cstring>#include<algorithm>#include<iostream>#include<time.h>#include<set>#include<stack>#include<vector>#include<map>#include<queue>#define pi acos(-1)#define maxn 111111#define maxm 11111#define INF 0x3F3F3F3F#define eps 1e-8#define pb push_back#define mem(a) memset(a,0,sizeof a)using namespace std;const long long mod = 1000000007;/**lyc**/void init() {}char s[111];int main() {    int n;    while(scanf("%d", &n) != EOF) {        scanf("%s", s);        int ans = 0;        for(int i = 0; i < n; i++) {            if(s[i] == '1') {                ans++;            }            else {                printf("%d", ans);                ans = 0;            }            //cout << ans <<endl;        }        printf("%d", ans);        printf("\n");    }    return 0;}

B.Five-In-a-Row

题目链接

题意:两人走五子棋，判断先手是否一步就能赢，枚举每一个五元组就好了

#include<cstdio>#include<cmath>#include<cstring>#include<algorithm>#include<iostream>#include<time.h>#include<set>#include<stack>#include<vector>#include<map>#include<queue>#define pi acos(-1)#define maxn 111111#define maxm 11111#define INF 0x3F3F3F3F#define eps 1e-8#define pb push_back#define mem(a) memset(a,0,sizeof a)using namespace std;const long long mod = 1000000007;/**lyc**/void init() {}char s[12][12];int a[12][12];int main() {    for(int i = 0; i < 10; i++) {        scanf("%s", s[i]);    }    for(int i = 1; i <= 10; i++) {        for(int j = 1; j <= 10; j++) {            if(s[i - 1][j - 1] == '.') a[i][j] = 0;            if(s[i - 1][j - 1] == 'X') a[i][j] = 1;            if(s[i - 1][j - 1] == 'O') a[i][j] = 2;        }    }    int fa = 0;    ///横的    int num0, num1;    for(int i = 1; i <= 10; i++) {        for(int j = 1; j <= 6; j++) {            num0 = 0; num1 = 0;            for(int k = 0; k < 5; k++) {                if(a[i][j + k] == 0) num0++;                if(a[i][j + k] == 1) num1++;            }            if(num1 == 4 && num0 == 1) {fa = 1;break;}        }        if(fa == 1) break;    }    ///竖的    for(int i = 1; i <= 10; i++) {        for(int j = 1; j <= 6; j++) {            num0 = 0; num1 = 0;            for(int k = 0; k < 5; k++) {                if(a[j + k][i] == 0) num0++;                if(a[j + k][i] == 1) num1++;            }            if(num1 == 4 && num0 == 1) {fa = 1;break;}        }        if(fa == 1) break;    }    ///右下    for(int i = 1; i <= 6; i++) {        for(int j = 1; j <= 6; j++) {            num0 = 0; num1 = 0;            for(int k = 0; k < 5; k++) {                if(a[j + k][i + k] == 0) num0++;                if(a[j + k][i + k] == 1) num1++;            }            if(num1 == 4 && num0 == 1) {fa = 1;break;}        }        if(fa == 1) break;    }    ///左下    for(int i = 5; i <= 10; i++) {        for(int j = 1; j <= 6; j++) {            num0 = 0; num1 = 0;            for(int k = 0; k < 5; k++) {                if(a[j + k][i - k] == 0) num0++;                if(a[j + k][i - k] == 1) num1++;            }            if(num1 == 4 && num0 == 1) {fa = 1;break;}        }        if(fa == 1) break;    }    if(fa == 1) printf("YES\n");    else printf("NO\n");    return 0;}

C.Multi-judge Solving

题目链接

题意: 首先有一个条件，如果要想刷难度是d的题目，之前一定要刷果难度>=d2 的题目, 然后给你n个难度的题目，和一个以之前刷过最高难度的题目，然后问你除了刷这些题目以外，还要额外刷多少题目。

思路:每次不断乘2就可以了。如果有断档,ans就+1

#include<cstdio>#include<cmath>#include<cstring>#include<algorithm>#include<iostream>#include<time.h>#include<set>#include<stack>#include<vector>#include<map>#include<queue>#define pi acos(-1)#define maxn 111111#define maxm 11111#define INF 0x3F3F3F3F#define eps 1e-8#define pb push_back#define mem(a) memset(a,0,sizeof a)using namespace std;const long long mod = 1000000007;/**lyc**/int n, k;int a[1111];int main() {    while(scanf("%d%d", &n, &k) != EOF) {        for(int i = 1; i <= n; i++) {            scanf("%d", &a[i]);        }        sort(a + 1, a + n + 1);        int ans = 0;        int l = 1;        while(l <= n) {            while(a[l] > 2 * k) {                ans++;                k *= 2;            }            k = max(k, a[l]);            l++;        }        printf("%d\n", ans);    }    return 0;}

D.Suitable Replacement

题目链接

题意:给你两个字符串s,t。然后s中有’?’即不确定的字母，可以随意变换。问s中最多能有多少个t，可以随意打乱字母顺序，所以最后就是一个统计一下字母个数,然后处理一下。最后填到s中，如果有多余就填个’a’好了。

#include<cstdio>#include<cmath>#include<cstring>#include<algorithm>#include<iostream>#include<time.h>#include<set>#include<stack>#include<vector>#include<map>#include<queue>#define pi acos(-1)#define maxn 111111#define maxm 11111#define INF 0x3F3F3F3F#define eps 1e-8#define pb push_back#define mem(a) memset(a,0,sizeof a)using namespace std;const long long mod = 1000000007;/**lyc**/void init() {}char s[1111111];char t[1111111];int nums[30];int numt[30];int anss[30];int main() {    scanf("%s%s", s, t);    int lens = strlen(s);    int lent = strlen(t);    memset(nums, 0, sizeof nums);    memset(numt, 0, sizeof numt);    int emp = 0;    for(int i = 0; i < lens; i++) {        if(s[i] == '?') emp++;        else nums[s[i] - 'a']++;    }    for(int i = 0; i < lent; i++) {        numt[t[i] - 'a']++;    }    int temp;    for(temp = 1; ; temp++) {        int res = 0;        for(int i = 0; i < 26; i++) {            res += max(numt[i] * temp - nums[i], 0);        }        if(res > emp) {            temp--;            break;        }    }    for(int i = 0; i < 26; i++) {        anss[i] = max(numt[i] * temp - nums[i], 0);    }   // for(int i = 0; i < 26; i++) {   //     cout << anss[i] << " ";   // }    int tt = 0;    for(int i = 0; i < lens; i++) {        if(s[i] == '?') {            while(anss[tt] == 0 && tt < 26) tt++;            if(tt == 26) {                s[i] = 'a';                continue;            }            s[i] = tt + 'a';            anss[tt]--;        }    }    printf("%s", s);    return 0;}