UVa1586

//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<stdio.h>
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 == '
'){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
", ans);
	}
	return 0;
}
#endif

#ifdef A2
#include<stdio.h>
#include<string.h>
#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<strlen(str); i++){
			if(str[i] == 'C') M = 12.01;
			if(str[i] == 'H') M = 1.008;
			if(str[i] == 'O') M = 16.00;
			if(str[i] == 'N') M = 14.01;
			if(i == strlen(str)-1 && str[i]>='C'){ ans += M; break;}
			if(sscanf(&str[i+1],"%d",&num) != 1) ans += M;
			else {
				i += bits(num);
				ans += M*num;
			}
		}
		printf("%.3f
", ans);
	}
	return 0;
}
#endif

#ifdef A3
#include<stdio.h>
#include<string.h>
#include<ctype.h> 
#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<len; i++){
			if(s[i] == 'C') n = 0;
			if(s[i] == 'H') n = 1;
			if(s[i] == 'O') n = 2;
			if(s[i] == 'N') n = 3;
			if(s[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
", 12.01*a[0]+1.008*a[1]+16.00*a[2]+14.01*a[3]);
	}
	return 0;
}
#endif

#ifdef A4
#include<stdio.h>
#include<string.h>
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
",j);
				}
				while(count--){a[n] += (s[i+count]-'1')*co(count);printf("%d ",count);}
			}
		}
	//输出
		printf("%.3f
",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
*/