UVa1586 - Molar mass

//UVa1586 - Molar mass
//给出一种由C, H, O, N 四种原子构成的分子式，求分子量
//#define A1 //无法处理换行问题(scanf)
//#define A2 //临界问题有BUG(sscanf)
#define A3 //只考虑两位即可AC，直接暴力
//#define A4 //考虑多位数有BUG

#ifdef A1
#include
int main(){
    //freopen("data.in","r",stdin);
    int T;
    scanf("%d",&T);
    getchar();
    while(T--){
        char ch, ch2 = 'A'; int num; float M, ans = 0;
        //无法处理换行问题
        //while((ch2=getchar()) != EOF){
        while(scanf("%c",&ch2) == 1){
            if(ch2 == '\n'){if(ch >= 'C')ans += M; break;}
            else ch = ch2;
            if(ch == 'C') M = 12.01;
            if(ch == 'H') M = 1.008;
            if(ch == 'O') M = 16.00;
            if(ch == 'N') M = 14.01;
            if(scanf("%d",&num) != 1) ans += M;
            else ans += M*num;
        }
        printf("%.3f\n", ans);
    }
    return 0;
}
#endif

#ifdef A2
#include
#include
#define maxn 20
int bits(int x){
    if(x == 0)return 1;
    int count = 0;
    while(x>0){ count++; x/=10;}
    return count;
}
int main(){
    freopen("data.in","r",stdin);
    int T;
    scanf("%d",&T);
    getchar();
    while(T--){
        int num; float M, ans = 0;
        char str[maxn];
        fgets(str,maxn,stdin);
        //scanf("%s",str);
        for(int i = 0; i='C'){ ans += M; break;}
            if(sscanf(&str[i+1],"%d",&num) != 1) ans += M;
            else {
                i += bits(num);
                ans += M*num;
            }
        }
        printf("%.3f\n", ans);
    }
    return 0;
}
#endif

#ifdef A3
#include
#include
#include 
#define maxn 100
int main(){
    //freopen("data.in","r",stdin);
    int T;
    scanf("%d",&T);
    while(T--){
        char s[maxn]; scanf("%s",s);
        int n, a[4], len = strlen(s);
        memset(a,0,sizeof(a));
        for(int i = 0; i='C')
            if(isdigit(s[i+1]) && isdigit(s[i+2])) a[n] += 10*(s[i+1]-'0')+(s[i+2]-'0');
            else a[n] += isdigit(s[i+1])? s[i+1]-'0': 1; 
        }
        printf("%.3f\n", 12.01*a[0]+1.008*a[1]+16.00*a[2]+14.01*a[3]);
    }
    return 0;
}
#endif

#ifdef A4
#include
#include
int co(int cot){int add42=1; while(--cot)add42 *= 10; return add42;}
int main(){
    freopen("data.in","r",stdin);
    int T;
    scanf("%d",&T);
    getchar();
    while(T--){
        char s[20];int a[4],n;
        memset(a,0,sizeof(a));
    //获取输入
        scanf("%s",s);
    //处理数据
        for(int i = 0; i < strlen(s); i++){
            if(s[i] > 57){//存储个数
                if(s[i] == 'C'){n=0;a[n]++;}
                if(s[i] == 'H'){n=1;a[n]++;}
                if(s[i] == 'O'){n=2;a[n]++;}
                if(s[i] == 'N'){n=3;a[n]++;}
            }else{
                int count = 0;//记录数字位数
                for(int j = 0; ; j++){
                    if(s[i+j] <= 57)count++;
                    else break;
                    printf("%d\n",j);
                }
                while(count--){a[n] += (s[i+count]-'1')*co(count);printf("%d ",count);}
            }
        }
    //输出
        printf("%.3f\n",12.01*a[0]+1.008*a[1]+16.00*a[2]+14.01*a[3]);
    }
    return 0;
}
#endif
/*测试数据:
4
C
C6H5OH
NH2CH2COOH
C12H22O11

12.010
94.108
75.070
342.296
*/