UVA - 1586 Molar mass?

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UVA - 1586 Molar mass?

题目大意：给一组化学式算出其质量
解题思路：一个一个算就好了
#include <cstdio>#include <cstring>#include <cstdlib>int main() {    char b[100];    char a[100];    int N;    scanf("%d", &N);    getchar();    double s;    while(N--) {        memset(a, 0, sizeof (a));        memset(b, 0, sizeof (b));        s = 0;        gets(a);        for ( int i = 0; i < strlen(a); i++) {            if (a[i] == 'C' ) {                if( a[i+1] > '9' || a[i+1] < '0' )                    s = s + 12.01;                else if (a[i+1] >= '0' && a[i+1] <= '9') {                    int k ;                    b[0] = a[i+1];                    for (  k = 1 ; k < strlen(a) - i; k++)                        if (a[i+1+k] < '0' || a[i+1+k] > '9') break;                        else {b[k] = a[i+1+k];b[k+1] = '\0';}                    s = s + 12.01*atoi(b);                    i = i + k;                    }            }            if (a[i] == 'H' ) {                if( a[i+1] > '9' || a[i+1] < '0' )                    s = s + 1.008;                else if (a[i+1] >= '0' && a[i+1] <= '9') {                    int k ;                    b[0] = a[i+1];                    for (  k = 1 ; k < strlen(a) - i; k++)                        if (a[i+1+k] < '0' || a[i+1+k] > '9') break;                        else {b[k] = a[i+1+k];b[k+1] = '\0';}                    s = s + 1.008*atoi(b);                    i = i  + k;                    }            }            if (a[i] == 'O' ) {                if( a[i+1] > '9' || a[i+1] < '0' )                    s = s + 16.00;                else if (a[i+1] >= '0' && a[i+1] <= '9') {                    int k ;                    b[0] = a[i+1];                    for (  k = 1 ; k < strlen(a) - i; k++)                        if (a[i+1+k] < '0' || a[i+1+k] > '9') break;                        else {b[k] = a[i+1+k]; b[k+1] = '\0';};                    s = s + 16.00*atoi(b);                    i=i + k;                    }            }            if (a[i] == 'N' ) {                if( a[i+1] > '9' || a[i+1] < '0' )                    s = s + 14.01;                else if (a[i+1] >= '0' && a[i+1] <= '9') {                    int k ;                    b[0] = a[i+1];                    for (  k = 1 ; k < strlen(a) - i; k++)                        if (a[i+1+k] < '0' || a[i+1+k] > '9') break;                        else  {b[k] = a[i+1+k]; b[k+1] = '\0';}                    s = s + 14.01*atoi(b);                    i=i + k;                    }            }         }        printf("%.3lf\n", s);    }}
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