Codeforces Round #419 (Div. 2)

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codeforces Round#419普通的一场div2.
codeforces816A
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题意：输入四个数，模拟时钟的走向，分针到60进位变成00，时针到24变成00，问最少经过多少步，使得其变成回文。
两种方法：第一种经过if，else判断，有几个坑，例如06:43等等，第二种方法直接while循环判断两个数字其中一个倒过来是否相同。我是用第一种判坑写的。

#include <bits/stdc++.h>#define INF 0x3f3f3f3f#define eps 1e-8#define MAXN (100000+10)#define MAXM (50000000)#define Si(a) scanf("%d", &a)#define Sl(a) scanf("%lld", &a)#define Sf(a) scanf("%lf", &a)#define Ss(a) scanf("%s", a)#define Sc(a) scanf(" %c", &a)#define Pi(a) printf("%d\n", (a))#define Pf(a) printf("%lf\n", (a))#define Pl(a) printf("%lld\n", (a))#define Ps(a) printf("%s\n", (a))#define MOD 1000000007#define LL long long#define lson  l, mid, o<<1#define rson mid+1, r, o<<1|1#define lowbits(x) (x & -x)using namespace std;int main(void){    int M, N, M1, N1;    char s[10];    scanf("%s", s);    M=s[0]-'0';    N=s[1]-'0';    M1=s[3]-'0';    N1=s[4]-'0';    int cnt=0;        if(N1==M&&M1==N)        {            printf("0\n");            return 0;        }        int m=N*10+M;        int n=M1*10+N1;        int k=M*10+N;        if(m>=n)        {            if(N<=5)            cnt=m-n;            else            {                if(M==0)                    cnt=60*(10-N)-(n-1);                else                    cnt=60*(10-N)-(n-2);            }        }        else        {            if(M*10+N<23)            {                if(N<5)                cnt=60-n+(k+1)/10+((k+1)%10)*10;                else                {                    if(M==0)                    {                        cnt=60*(10-N)-(n-1);                    }                    else                        cnt=60*(10-N)-(n-2);                }            }            else                cnt=60-n;        }        printf("%d\n", cnt);    return 0;}

第二题题意：给你n个区间，然后有k个询问区间，问这k个区间中每个区间在n个区间重叠部分大于m的数有几个…（有点智障说不清楚）以为是线段树，其实用个辅助数组a[i]表示第i个元素以后全部加a[i]..
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代码如下：

#include <bits/stdc++.h>#define INF 0x3f3f3f3f#define eps 1e-8#define MAXN (100000+10)#define MAXM (50000000)#define Si(a) scanf("%d", &a)#define Sl(a) scanf("%lld", &a)#define Sf(a) scanf("%lf", &a)#define Ss(a) scanf("%s", a)#define Sc(a) scanf(" %c", &a)#define Pi(a) printf("%d\n", (a))#define Pf(a) printf("%lf\n", (a))#define Pl(a) printf("%lld\n", (a))#define Ps(a) printf("%s\n", (a))#define MOD 1000000007#define LL long long#define lson  l, mid, o<<1#define rson mid+1, r, o<<1|1#define lowbits(x) (x & -x)using namespace std;int n, k, q, a[200100], b[200100];int main(void){    while(scanf("%d%d%d", &n, &k, &q)!=EOF)    {        int x, y;        memset(a, 0, sizeof(a));        memset(b, 0, sizeof(b));        for(int i=1; i<=n; i++)        {            scanf("%d%d", &x,&y);            a[x]++;            a[y+1]--;        }        for(int i=1; i<=200100; i++)        {            a[i]+=a[i-1];            b[i]=b[i-1]+(a[i]>=k);        }        for(int i=1; i<=q; i++)        {            scanf("%d%d", &x, &y);            printf("%d\n", b[y]-b[x-1]);        }    }    return 0;}

第三题题意：初始为0的二维数组，每项步骤为行或者列加一，求最少步骤。很简单的模拟题，只需要判断行和列的大小再进行相应模拟。
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代码如下：

#include <bits/stdc++.h>#define INF 0x3f3f3f3f#define eps 1e-8#define MAXN (100000+10)#define MAXM (50000000)#define Si(a) scanf("%d", &a)#define Sl(a) scanf("%lld", &a)#define Sf(a) scanf("%lf", &a)#define Ss(a) scanf("%s", a)#define Sc(a) scanf(" %c", &a)#define Pi(a) printf("%d\n", (a))#define Pf(a) printf("%lf\n", (a))#define Pl(a) printf("%lld\n", (a))#define Ps(a) printf("%s\n", (a))#define MOD 1000000007#define LL long long#define lson  l, mid, o<<1#define rson mid+1, r, o<<1|1#define lowbits(x) (x & -x)using namespace std;int M, N, a[200][200];struct node{    string s;    int x, y;};vector<node> v;int cnt;void up1(){    for(int i=1; i<=M; i++)        {            int Min=INF;            for(int j=1; j<=N; j++)            Min=min(Min, a[i][j]);            cnt+=Min;            if(Min==0)continue;            node n;            for(int j=1; j<=N; j++)                a[i][j]-=Min;            n.s="row", n.x=i, n.y=Min;            v.push_back(n);        }}void up2(){    for(int i=1; i<=N; i++)        {            int Min=INF;            for(int j=1; j<=M; j++)            Min=min(Min, a[j][i]);            cnt+=Min;            if(Min==0)continue;            node n;            for(int j=1; j<=M; j++)                a[j][i]-=Min;            n.s="col", n.x=i, n.y=Min;            v.push_back(n);        }}int main(void){    while(scanf("%d%d",  &M, &N)!=EOF)    {        cnt=0;        memset(a, 0, sizeof(a));        for(int i=1; i<=M; i++)            for(int j=1; j<=N; j++)            scanf("%d", &a[i][j]);        if(M<=N)up1(),up2();        else up2(), up1();        bool flag=true;        for(int i=1; i<=M; i++)            for(int j=1; j<=N; j++)            if(a[i][j])flag=false;        if(flag)        {            printf("%d\n", cnt);            for(int i=1; i<=v.size(); i++)            {                for(int j=1; j<=v[i-1].y; j++)                cout<<v[i-1].s<<" "<<v[i-1].x<<endl;            }        }        else cout<<"-1"<<endl;    }    return 0;}

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