数据结构作业-3.4累加的递归实现

#include <stdio.h> //计算1到paraN的累加和 int addTo(int paraN){ int tempSum; printf(" entering addTo(%d)\r\n",paraN); if(paraN<=0){ printf("return 0\r\n"); return 0; }else{ tempSum = addTo(paraN - 1) + paraN; printf("return %d\r\n",tempSum); return tempSum; } } //极简版递归求和函数 int clearAddTo(int paraN) { if (paraN <= 0) { return 0; } else { return clearAddTo(paraN - 1) + paraN; } } //测试 void addToTest() { int n, sum; printf("---- addToTest begins. ----\r\n"); n = 5; sum = addTo(n); printf("\r\n0 adds to %d gets %d.\r\n", n, sum); n = 1; sum = addTo(n); printf("\r\n0 adds to %d gets %d.\r\n", n, sum); n = -1; sum = addTo(n); printf("\r\n0 adds to %d gets %d.\r\n", n, sum); printf("---- addToTest ends. ----\r\n"); } void main() { addToTest(); }