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app.c for zyy
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#include <stdio.h> | |
int main(int argc, char const *argv[]) | |
{ | |
int count; | |
double input, output; | |
// 0. prepare count | |
scanf("%d", &count); | |
printf("count: %d\n", count); | |
int levels[count + 1]; | |
double ratios[count + 1]; | |
count += 1; | |
// 1. prepare levels and ratio | |
levels[0] = 0; | |
ratios[0] = 0; | |
for (int i = 1; i < count; i++) { | |
scanf("%d %lf", &levels[i], &ratios[i]); | |
printf("%d %lf\n", levels[i], ratios[i]); | |
} | |
scanf("%lf", &input); | |
printf("input: %.2f\n", input); | |
// 2. process | |
for (int i = 1; i < count - 1; i++) { | |
// reach the level ceiling, break | |
if (input < levels[i]) { | |
output += (input - levels[i - 1]) * ratios[i]; | |
printf("stage 1 %.2f\n", output); | |
break; | |
} | |
// accumulate current level | |
output += (levels[i] - levels[i - 1]) * ratios[i]; | |
printf("stage 2 %.2f\n", output); | |
} | |
// last level | |
if (input > levels[count - 2]) { | |
output += (input - levels[count - 1 - 1]) * ratios[count - 1]; | |
printf("stage 3 %.2f\n", output); | |
} | |
// 3. output | |
printf("%.2f", output); | |
return 0; | |
} | |
// gcc app.c | |
// echo '3\n1000 0.1\n2000 0.2\n-1 0.3\n2500' | ./a.out | |
// echo '6\n100000 0.1\n200000 0.075\n400000 0.05\n600000 0.03\n1000000 0.015\n-1 0.01\n150000' | ./a.out |
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