X is an ndim positively curved manifold and Y is a totally geodesic submanifold of codimension 1.Then cutting along Y we get ndim positively curved manifolds without boundary,by soul theorem these manifolds should be homeomorphic to R^n.Am I right?If not,please give counterexamples.

If I understood your question correctly, the answer is yes. More precisely, the following statement should hold:
This follows, as the OP suggests, from the Soul Argument of CheegerGromoll, extended to Alexandrov spaces by Perelman (see, e.g., Section 6 of Perelman's notes). As mentioned in the comments, CheegerGromoll's version of the argument actually suffices to get the conclusion. A few details: denote by $C_1$ and $C_2$ the closure of the two connected components of $X\setminus Y$. These are positively curved compact Alexandrov spaces with boundary $Y$. On each of them, since the curvature is positive, the distance function to the boundary is concave. Therefore, the set of points at maximal distance (the soul) consists of a unique point. This implies that each $C_i$ is homeomorphic to a disk, hence $X=C_1\cup_{Y} C_2$ is a twisted sphere. edit (to answer GB's comment): As discussed above, if $C$ a compact Alexandrov space with curvatures $\geq k>0$, then the soul $S=\{p\}$ is a point. Moreover, according to Perelman, the pairs $(C,\partial C)$ and $(\overline K(\Sigma_S),\Sigma_S)$ are homeomorphic (see 6.2 for proof), where $\Sigma_S$ is the space of directions at the soul and $\overline K(\Sigma_S)$ is the closure of the topological cone over $\Sigma_S$, i.e., the join of $\Sigma_S$ and a point. If $C$ is a manifold, the space of directions are spheres, so $(\overline K(\Sigma_S),\Sigma_S)$ is simply a pair $(D,\partial D)$, where $D$ is a disk. 

