第三章数组和字符串下

第三章数组和字符串下(๑•ᴗ•๑)

Uva1368 DNA Consensus String

//一道典型的水题。。。。。
#include
#include
#include

using namespace std;
int it[5];
char its[5]="ACGT";
int main(){
    int T,n,m;
    scanf("%d",&T);
    while(T--){
        int errors = 0;
        int num_A=0,num_C=0,num_G=0,num_T=0;
        scanf("%d%d",&n,&m);
        char a[n][m];
        char b[m];
        for (int i = 0; i < n; i++)
            scanf("%s",a[i]);
        for (int i = 0; i < m; i++){
            memset(it, 0, sizeof(it));
            for (int j = 0; j < n; j++)
                for (int q = 0; q < 4; q++)
                    if (its[q] == a[j][i]) {it[q]++;break;}
            if (it[0] >= max(max(it[1],it[2]),it[3]) ) b[i] = its[0];
            else if(it[1] >= max(it[2],it[3]) ) b[i] = its[1];
            else if(it[2] >= it[3] ) b[i] = its[2];
            else b[i] = its[3];
            for (int j = 0; j < n; j++)
                if (a[j][i] != b[i]) errors++;
            cout << b[i];
        }
        cout << endl << errors << endl;
    }
    return 0;
}

UVa202 Repeating Decimals

//找一个莫名其妙的bug找了半天，看来还是刚开始思路不清晰；(＞﹏＜)
#include
#include
#include
using namespace std;
int yu[4000];
int chu[4000];
int main(){
    int n,m,i,k,l;
    while(~scanf("%d%d",&n,&m)){
        bool flag = 1;
        memset(yu, 0, sizeof(yu));
        memset(chu, 0, sizeof(chu));
        chu[0] = n/m;
        yu[0] = n%m;
        printf("%d/%d = %d.",n,m,chu[0]);
        for (i = 1;i < 4000; i++){
            yu[i] = yu[i-1]*10 % m;
            chu[i] = yu[i-1]*10 / m;
            if(yu[i] == 0){break;}
        }
        //cout  <
        for(k = 1; flag; k++)
        for(l = k+1; l <= k+m; l++){
            if(chu[k] == chu[l] && yu[k] == yu[l]){
                flag = 0;
                break;
            }
        }
        if(k==1)cout <<'(';
        for (int j = 1;; j++){
            if (j == l) {cout << ')'<break;}
            else if(j == k-1) {cout << '(';}
            if(j <= 50)cout <else {cout << "...)" <break;}
        }
        printf("   %d = number of digits in repeating cycle\n\n",l-k+1);
    }
    return 0;
}

UVa 10340 All in All

//水题
#include
#include
#include
using namespace std;

int main(){
    string s,t;
    int i,j;
    while(cin >> s >> t){
        for(i = 0,j = 0; s[i] && t[j];){
            if (s[i]==t[j]){i++;j++;}
            else j++;
        }
        if(s[i])printf("No\n");
        else printf("Yes\n");
    }
    return 0;
}

UVa 1587 Box

//抄来的，这个头文件好清爽。。。
#include
using namespace std;
struct face{
    int x,y;
}a[6];

bool cmp(face a, face b){
    if(a.x == b.x) return a.y > b.y;
    else return a.x > b.x;
}

bool judge(){
    if(memcmp(a,a+1,sizeof(face)) || memcmp(a+2,a+3,sizeof(face)) ||
        memcmp(a+4,a+5,sizeof(face))) return 0;//成对出现
    if(a[1].x!=a[2].x||a[1].y!=a[4].x||a[2].y!=a[4].y)  return false;//相连边相等
  return true;
}
int main(){
    while(cin>>a[0].x>>a[0].y>>a[1].x>>a[1].y>>a[2].x>>a[2].y>>a[3].x>>a[3].y>>a[4].x>>a[4].y>>a[5].x>>a[5].y){
        for(int i=0;i<6;i++)
            if(a[i].x<a[i].y)
                swap(a[i].x,a[i].y);
        sort(a,a+6,cmp);
        printf("%s\n",judge()?"POSSIBLE":"IMPOSSIBLE");
    }
    return 0;
}

UVa 1588 Kickdown

#include
#include
#include
#include
#include
using namespace std;

int main(){
    string a,b;
    while(cin >> a >> b){
        int n[300]={0},i;
        int n2[300]={0};
        bool flag = 1;
        int maxn = 0;
        for (i = 0; flag && a[i];i++){
            for (int j = 0;b[j]; j++){
            if(!a[i+j]&&b[j]){flag = 0;break;}
            n[i]++;
            if(a[i+j]+b[j]>'1'+'2'){n[i]=0;break;}
            //cout << a[i+j] + b[j] << ":" << n[i]<
            }
            maxn=max(maxn,n[i]);
        }
      //对调主板。。。
        flag = 1;
        for (i = 0; flag && b[i];i++){
            for (int j = 0;a[j]; j++){
            if(!b[i+j]&&a[j]){flag = 0;break;}
            n2[i]++;
            if(b[i+j]+a[j]>'1'+'2'){n2[i]=0;break;}
            //cout << a[i+j] + b[j] << ":" << n[i]<
            }
            maxn=max(maxn,n2[i]);
        }
        //cout << maxn << endl;
        //if(maxn == 0)maxn == 1;
        cout << a.length()+b.length()-maxn <return 0;
}

UVa 11809 Floating-Point Numbers

//这题也是借鉴来的
#include
#include
#include
#include
using namespace std;
long long E[20][40];//存放指数次数
double M[20][40];//存放<10的部分
int main(){
    int e=1;
    for(int i = 0 ; i <=9; i++)for(int j = 1; j <= 30; j++){
        double m = 1 - pow(0.5,i+1),e = pow(2, j) - 1;
        double t = log10(m) + e*log10(2);//取对数防止爆
        E[i][j] = (long long)t, M[i][j] = pow(10,t-E[i][j]);
    }
    string it;
    while(cin >> it && it!="0e0"){
        double Man=0;
        int Expo=0;
        int i=0,j;
        //读取，精度有损失
        while(it[i] != '.')
            Man = Man*10 + it[i++]-'0';
        j=i+1;
        while(it[j] != 'e')
            Man += (double)pow(0.1,j-i)*(it[j++]-'0');
        while(it[++j])
            Expo= Expo*10+it[j]-'0';
        while(Man<1)Man*=10,Expo-=1;
        //查表
        for(int i = 0; i <= 9; i++)for(int j = 1; j <= 30; j++){
            if(Expo == E[i][j])
            if(fabs(Man - M[i][j]) < 1e-4)//有误差
            cout << i << " " << j << endl;
        }
    }
    return 0;
}