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#include <stdio.h>
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#include <time.h>
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int main() {
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long timea, timeb, timec;//clock_t timea;
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long a = 1LL, b = 100000000LL;
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long long result1 = 0LL, result2 = 0LL;
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timea = clock();
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for (long long i = a; i <= b; i++) {
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result1 += i;
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}
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timeb = clock();
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printf("累加结果: %lld, 所用时间: %lf 毫秒\n", result1, (double)(timeb - timea) / CLOCKS_PER_SEC * 1000);
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result2 = (a + b) * (b - a + 1) / 2;
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timec = clock();
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printf("高斯结果: %lld, 所用时间: %lf 毫秒\n", result2, (double)(timec - timeb) / CLOCKS_PER_SEC * 1000);
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return 0;
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}
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